Arithmetic–geometric mean

To find the arithmetic–geometric mean of a₀ = 24 and g₀ = 6, iterate as follows:

$${\displaystyle {\begin{array}{rcccl}a_{1}&=&{\tfrac {1}{2}}(24+6)&=&15\\g_{1}&=&{\sqrt {24\cdot 6}}&=&12\\a_{2}&=&{\tfrac {1}{2}}(15+12)&=&13.5\\g_{2}&=&{\sqrt {15\cdot 12}}&=&13.416\ 407\ 8649\dots \\&&\vdots &&\end{array}}}$$

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 13.4581714817256154207668131569743992430538388544.[2]

The first algorithm based on this sequence pair appeared in the works of Lagrange. Its properties were further analyzed by Gauss.[1]

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 n > 0, (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; it is also between x and y.

If r ≥ 0, then M(rx,ry) = r M(x,y).

There is an integral-form expression for M(x,y):[3]

$${\displaystyle {\begin{aligned}M(x,y)&={\frac {\pi }{2}}\left(\int _{0}^{\frac {\pi }{2}}{\frac {d\theta }{\sqrt {x^{2}\cos ^{2}\theta +y^{2}\sin ^{2}\theta }}}\right)^{-1}\\&=\pi \left(\int _{0}^{\infty }{\frac {dt}{\sqrt {t(t+x^{2})(t+y^{2})}}}\right)^{-1}\\&={\frac {\pi }{4}}\cdot {\frac {x+y}{K\left({\frac {x-y}{x+y}}\right)}}\end{aligned}}}$$

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

$${\displaystyle K(k)=\int _{0}^{\frac {\pi }{2}}{\frac {d\theta }{\sqrt {1-k^{2}\sin ^{2}(\theta )}}}}$$

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.[4]

The arithmetic–geometric mean is connected to the Jacobi theta function ${\displaystyle \theta _{3}}$ by[5]

$${\displaystyle M(1,x)=\theta _{3}^{-2}\left(\exp \left(-\pi {\frac {M(1,x)}{M\left(1,{\sqrt {1-x^{2}}}\right)}}\right)\right)=\left(\sum _{n\in \mathbb {Z} }\exp \left(-n^{2}\pi {\frac {M(1,x)}{M\left(1,{\sqrt {1-x^{2}}}\right)}}\right)\right)^{-2},}$$

which upon setting ${\displaystyle x=1/{\sqrt {2}}}$ gives

$${\displaystyle M(1,1/{\sqrt {2}})=\left(\sum _{n\in \mathbb {Z} }e^{-n^{2}\pi }\right)^{-2}.}$$

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

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

In 1799, Gauss proved[note 1] 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 proven transcendental by Theodor Schneider.[note 2][6][7] The set ${\displaystyle \{\pi ,M(1,1/{\sqrt {2}})\}}$ is algebraically independent over ${\displaystyle \mathbb {Q} }$,[8][9] but the set ${\displaystyle \{\pi ,M(1,1/{\sqrt {2}}),M'(1,1/{\sqrt {2}})\}}$ (where the prime denotes the derivative with respect to the second variable) is not algebraically independent over ${\displaystyle \mathbb {Q} }$. In fact,[10]

$${\displaystyle \pi =2{\sqrt {2}}{\frac {M^{3}(1,1/{\sqrt {2}})}{M'(1,1/{\sqrt {2}})}}.}$$

The geometric–harmonic mean can be calculated by an analogous method, using sequences of geometric and harmonic means. One finds that GH(x,y) = 1/M(1/x, 1/y) = xy/M(x,y).[11] 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,[12] and Jacobi elliptic functions.[13]

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

$${\displaystyle g_{n}\leq a_{n}}$$

and thus

$${\displaystyle g_{n+1}={\sqrt {g_{n}\cdot a_{n}}}\geq {\sqrt {g_{n}\cdot g_{n}}}=g_{n}}$$

that is, the sequence g_n 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 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 g such that:

$${\displaystyle \lim _{n\to \infty }g_{n}=g}$$

However, we can also see that:

$${\displaystyle a_{n}={\frac {g_{n+1}^{2}}{g_{n}}}}$$

and so:

$${\displaystyle \lim _{n\to \infty }a_{n}=\lim _{n\to \infty }{\frac {g_{n+1}^{2}}{g_{n}}}={\frac {g^{2}}{g}}=g}$$

Q.E.D.

This proof is given by Gauss.[1] 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 ${\displaystyle \theta '}$, where

$${\displaystyle \sin \theta ={\frac {2x\sin \theta '}{(x+y)+(x-y)\sin ^{2}\theta '}},}$$

gives

$${\displaystyle {\begin{aligned}I(x,y)&=\int _{0}^{\pi /2}{\frac {d\theta '}{\sqrt {{\bigl (}{\frac {1}{2}}(x+y){\bigr )}^{2}\cos ^{2}\theta '+{\bigl (}{\sqrt {xy}}{\bigr )}^{2}\sin ^{2}\theta '}}}\\&=I{\bigl (}{\tfrac {1}{2}}(x+y),{\sqrt {xy}}{\bigr )}.\end{aligned}}}$$

Thus, we have

$${\displaystyle {\begin{aligned}I(x,y)&=I(a_{1},g_{1})=I(a_{2},g_{2})=\cdots \\&=I{\bigl (}M(x,y),M(x,y){\bigr )}=\pi /{\bigr (}2M(x,y){\bigl )}.\end{aligned}}}$$

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

Finally, we obtain the desired result

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

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