Arithmetic–geometric mean

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 g₀ ≤ g₁ ≤ g₂ ≤ ...; the arithmetic means are a decreasing sequence a₀ ≥ a₁ ≥ a₂ ≥ ...; and g_n ≤ M(x, y) ≤ a_n 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 M(rx, ry) = r M(x, y).

There is an integral-form expression for M(x, y):$${\displaystyle \begin{array}{cc}M(x,y)& =\frac{\pi }{2}{\left({\int }_0^{\frac{\pi }{2}}\frac{d\theta }{\sqrt{x^2{\cos }^2\theta +y^2{\sin }^2\theta }}\right)}^{-1}\\ & =\pi {\left({\int }_0^{\mathrm{\infty }}\frac{dt}{\sqrt{t(t+x^2)(t+y^2)}}\right)}^{-1}\\ & =\frac{\pi }{4}\cdot \frac{x+y}{K\left(\frac{x-y}{x+y}\right)}\end{array}}$$where K(k) is the complete elliptic integral of the first kind:$${\displaystyle K(k)={\int }_0^{\frac{\pi }{2}}\frac{d\theta }{\sqrt{1-k^2{\sin }^2\theta }}}$$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 ${\displaystyle {\theta }_3}$ by$${\displaystyle \begin{array}{cc}M(1,x)& ={\theta }_3^{-2}\left(\exp \left(-\pi \frac{M(1,x)}{M\left(1,\sqrt{1-x^2}\right)}\right)\right)\\ & ={\left[\sum _{n\in \mathbb{Z}}\exp \left(-n^2\pi \frac{M(1,x)}{M\left(1,\sqrt{1-x^2}\right)}\right)\right]}^{-2},\end{array}}$$which upon setting ${\displaystyle x=1/\sqrt{2}}$ gives$${\displaystyle M(1,1/\sqrt{2})={\left(\sum _{n\in \mathbb{Z}}{e}^{-n^2\pi }\right)}^{-2}.}$$

Related concepts

The reciprocal of the arithmetic–geometric mean of 1 and the square root of 2 is Gauss's constant.$${\displaystyle \frac{1}{M(1,\sqrt{2})}=G=0.8346268\dots }$$In 1799, Gauss proved that$${\displaystyle M(1,\sqrt{2})=\frac{\pi }{\varpi }}$$where ${\displaystyle \varpi }$ is the lemniscate constant.

In 1941, ${\displaystyle M(1,\sqrt{2})}$ (and hence ${\displaystyle G}$) was proved transcendental by Theodor Schneider. The set ${\displaystyle \{\pi ,M(1,1/\sqrt{2})\}}$ is algebraically independent over ${\displaystyle \mathbb{Q}}$, but the set ${\displaystyle \{\pi ,M(1,1/\sqrt{2}),M^{\prime }(1,1/\sqrt{2})\}}$ (where the prime denotes the derivative with respect to the second variable) is not algebraically independent over ${\displaystyle \mathbb{Q}}$. In fact,$${\displaystyle \pi =2\sqrt{2}\frac{M^3(1,1/\sqrt{2})}{M^{\prime }(1,1/\sqrt{2})}.}$$The geometric–harmonic mean GH can be calculated using analogous sequences of geometric and harmonic means, and in fact GH(x, y) = 1/M(1/x, 1/y) = xy/M(x, y). 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 that$${\displaystyle g_n\le a_n}$$and thus$${\displaystyle g_{n+1}=\sqrt{g_n\cdot a_n}\ge \sqrt{g_n\cdot g_n}=g_n}$$that is, the sequence g_n 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:$${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{g}_n=g}$$However, we can also see that:$${\displaystyle a_n=\frac{g_{n+1}^2}{g_n}}$$ and so: $${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n=\underset{n\to \mathrm{\infty }}{lim}\frac{g_{n+1}^2}{g_n}=\frac{g^2}{g}=g}$$

Q.E.D.

Proof of the integral-form expression

This proof is given by Gauss. Let

$${\displaystyle I(x,y)={\int }_0^{\pi /2}\frac{d\theta }{\sqrt{x^2{\cos }^2\theta +y^2{\sin }^2\theta }},}$$

Changing the variable of integration to ${\displaystyle {\theta }^{\prime }}$, where

$${\displaystyle \begin{array}{cc}\sin \theta & =\frac{2x\sin {\theta }^{\prime }}{(x+y)+(x-y){\sin }^2{\theta }^{\prime }}\\ \Rightarrow d(\sin \theta )& =d\left(\frac{2x\sin {\theta }^{\prime }}{(x+y)+(x-y){\sin }^2{\theta }^{\prime }}\right)\\ \Rightarrow \cos \theta d\theta & =2x\frac{(x+y)-(x-y){\sin }^2{\theta }^{\prime }}{((x+y)+(x-y){\sin }^2{\theta }^{\prime }{)}^2}\cos {\theta }^{\prime }d{\theta }^{\prime }\end{array}}$$

$${\displaystyle \begin{array}{cc}\cos \theta & =\frac{\sqrt{(x+y{)}^2-2(x^2+y^2){\sin }^2{\theta }^{\prime }+(x-y{)}^2{\sin }^4{\theta }^{\prime }}}{(x+y)+(x-y){\sin }^2{\theta }^{\prime }}\\ & =\frac{\cos {\theta }^{\prime }\sqrt{(x-y{)}^2{\cos }^2{\theta }^{\prime }+4xy}}{(x+y)+(x-y){\sin }^2{\theta }^{\prime }}\\ & =\frac{\cos {\theta }^{\prime }\sqrt{(x+y{)}^2{\cos }^2{\theta }^{\prime }+4xy{\sin }^2{\theta }^{\prime }}}{(x+y)+(x-y){\sin }^2{\theta }^{\prime }},\end{array}}$$

$${\displaystyle \begin{array}{cc}\Rightarrow \cos \theta d\theta & =\frac{\cos {\theta }^{\prime }\sqrt{(x+y{)}^2{\cos }^2{\theta }^{\prime }+4xy{\sin }^2{\theta }^{\prime }}}{(x+y)+(x-y){\sin }^2{\theta }^{\prime }}d\theta \\ & =2x\frac{(x+y)-(x-y){\sin }^2{\theta }^{\prime }}{((x+y)+(x-y){\sin }^2{\theta }^{\prime }{)}^2}\cos {\theta }^{\prime }d{\theta }^{\prime },\end{array}}$$

$${\displaystyle \Rightarrow d\theta =\frac{x((x+y)-(x-y){\sin }^2{\theta }^{\prime })}{((x+y)+(x-y){\sin }^2{\theta }^{\prime })}\frac{2d{\theta }^{\prime }}{\sqrt{(x+y{)}^2{\cos }^2{\theta }^{\prime }+4xy{\sin }^2{\theta }^{\prime }}},}$$

$${\displaystyle \begin{array}{cc}\sqrt{x^2{\cos }^2\theta +y^2{\sin }^2\theta }& =\frac{\sqrt{x^2\left[(x+y{)}^2-2(x^2+y^2){\sin }^2{\theta }^{\prime }+(x-y{)}^2{\sin }^4{\theta }^{\prime }\right]+4x^2{y}^2{\sin }^2{\theta }^{\prime }}}{((x+y)+(x-y){\sin }^2{\theta }^{\prime })}\\ & =\frac{x\left((x+y)-(x-y){\sin }^2{\theta }^{\prime }\right)}{\left((x+y)+(x-y){\sin }^2{\theta }^{\prime }\right)}\end{array}}$$

This yields $${\displaystyle \begin{array}{cc}\frac{d\theta }{\sqrt{x^2{\cos }^2\theta +y^2{\sin }^2\theta }}& =\frac{2d{\theta }^{\prime }}{\sqrt{(x+y{)}^2{\cos }^2{\theta }^{\prime }+4xy{\sin }^2{\theta }^{\prime }}}\\ & =\frac{d{\theta }^{\prime }}{\sqrt{{\left(\frac{x+y}{2}\right)}^2{\cos }^2{\theta }^{\prime }+{\left(\sqrt{xy}\right)}^2{\sin }^2{\theta }^{\prime }}},\end{array}}$$

gives

$${\displaystyle \begin{array}{cc}I(x,y)& ={\int }_0^{\pi /2}\frac{d{\theta }^{\prime }}{\sqrt{{\left(\frac{x+y}{2}\right)}^2{\cos }^2{\theta }^{\prime }+{\left(\sqrt{xy}\right)}^2{\sin }^2{\theta }^{\prime }}}\\ & =I\left(\frac{x+y}{2},\sqrt{xy}\right).\end{array}}$$

Thus, we have

$${\displaystyle \begin{array}{cc}I(x,y)& =I(a_1,g_1)=I(a_2,g_2)=\cdots \\ & =I(M(x,y),M(x,y))=\frac{\pi }{2M(x,y)}.\end{array}}$$ The last equality comes from observing that ${\displaystyle I(z,z)=\pi /(2z)}$.

Finally, we obtain the desired result

$${\displaystyle M(x,y)=\frac{\pi }{2I(x,y)}.}$$

Applications

The number π

According to the Gauss–Legendre algorithm,

$${\displaystyle \pi =\frac{4M(1,1/\sqrt{2}{)}^2}{1-{\displaystyle \sum _{j=1}^{\mathrm{\infty }}{2}^{j+1}{c}_j^2}},}$$

where

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

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

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

Complete elliptic integral K(sinα)

Taking ${\displaystyle a_0=1}$ and ${\displaystyle g_0=\cos \alpha }$ yields the AGM

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

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

$${\displaystyle K(k)={\int }_0^{\pi /2}\frac{d\theta }{\sqrt{1-k^2{\sin }^2\theta }}.}$$

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

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.