arithmetic–geometric mean

To find the arithmetic–geometric mean of {{math|a₀ {{=}} 24}} and {{math|g₀ {{=}} 6}}, iterate as follows:$$\backslash begin\{array\}\{rcccl\}$$

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 {{math|a_n}} and {{math|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 {{val|13.4581714817256154207668131569743992430538388544}}.[http://www.wolframalpha.com/input/?i=agm%2824%2C+6%29 agm(24, 6)] at Wolfram Alpha

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

Both the geometric mean and arithmetic mean of two positive numbers {{mvar|x}} and {{mvar|y}} are between the two numbers. (They are strictly between when {{math|x ≠ y}}.) The geometric mean of two positive numbers is never greater than the arithmetic mean.{{cite book |last=Bullen |first=P. S. |contribution=The Arithmetic, Geometric and Harmonic Means |date=2003 |url=http://link.springer.com/10.1007/978-94-017-0399-4_2 |title=Handbook of Means and Their Inequalities |pages=60–174 |access-date=2023-12-11 |place=Dordrecht |publisher=Springer Netherlands |language=en |doi=10.1007/978-94-017-0399-4_2 |isbn=978-90-481-6383-0}} So the geometric means are an increasing sequence {{math|g{{sub|0}} ≤ g{{sub|1}} ≤ g{{sub|2}} ≤ ...}}; the arithmetic means are a decreasing sequence {{math|a{{sub|0}} ≥ a{{sub|1}} ≥ a{{sub|2}} ≥ ...}}; and {{math|g_n ≤ M(x, y) ≤ a_n}} for any {{mvar|n}}. These are strict inequalities if {{math|x ≠ y}}.

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

If {{math|r ≥ 0}} then {{math|M(rx, ry) {{=}} r M(x, y)}}.

There is an integral-form expression for {{math|M(x, y)}}:{{dlmf|first1=B. C.|last1=Carson|id=19.8.i|title=Elliptic Integrals|mode=cs1}}$$\backslash begin\{align\}$$

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 {{math|K(k)}} is the complete elliptic integral of the first kind:$$K(k)\; =\; \backslash int\_0^\backslash frac\{\backslash pi\}\{2\}\backslash frac\{d\backslash theta\}\{\backslash sqrt\{1\; -\; k^2\backslash sin^2\backslash theta\}\}$$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.{{cite book |author-first=Hercules G. |author-last=Dimopoulos |title=Analog Electronic Filters: Theory, Design and Synthesis |url=https://books.google.com/books?id=6W1eX4QwtyYC&pg=PA147 |year=2011 |publisher=Springer |isbn=978-94-007-2189-0 |pages=147–155 }}

The arithmetic–geometric mean is connected to the Jacobi theta function $\backslash theta\_3$ by{{Cite book |last1=Borwein |first1=Jonathan M. |last2=Borwein| first2=Peter B. |title=Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity |publisher=Wiley-Interscience |year=1987 |edition=First |isbn=0-471-83138-7}} pages 35, 40$$M(1,x)=\backslash theta\_3^\{-2\}\backslash left(\backslash exp\; \backslash left(-\backslash pi\; \backslash frac\{M(1,x)\}\{M\backslash left(1,\backslash sqrt\{1-x^2\}\backslash right)\}\backslash right)\backslash right)=\backslash left(\backslash sum\_\{n\backslash in\backslash mathbb\{Z\}\}\backslash exp\; \backslash left(-n^2\; \backslash pi\; \backslash frac\{M(1,x)\}\{M\backslash left(1,\backslash sqrt\{1-x^2\}\backslash right)\}\backslash right)\backslash right)^\{-2\},$$which upon setting $x=1/\backslash sqrt\{2\}$ gives$$M(1,1/\backslash sqrt\{2\})=\backslash left(\backslash sum\_\{n\backslash in\backslash mathbb\{Z\}\}e^\{-n^2\backslash pi\}\backslash right)^\{-2\}.$$

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

In 1941, $M(1,\backslash sqrt\{2\})$ (and hence $G$) was proved transcendental by Theodor Schneider.In particular, he proved that the beta function $\backslash Beta\; (a,b)$ is transcendental for all $a,b\backslash in\backslash mathbb\{Q\}\backslash setminus\backslash mathbb\{Z\}$ such that $a+b\backslash notin\; \backslash mathbb\{Z\}\_0^-$. The fact that $M(1,\backslash sqrt\{2\})$ is transcendental follows from $M(1,\backslash sqrt\{2\})=\backslash tfrac\{1\}\{2\}\backslash Beta\; \backslash left(\backslash tfrac\{1\}\{2\},\backslash tfrac\{3\}\{4\}\backslash right).${{cite journal |first=Theodor |last=Schneider |url=https://www.deepdyve.com/lp/de-gruyter/zur-theorie-der-abelschen-funktionen-und-integrale-mn0U50bvkB |title=Zur Theorie der Abelschen Funktionen und Integrale |year=1941 |journal=Journal für die reine und angewandte Mathematik |volume=183 |number=19 |pages=110–128 |doi=10.1515/crll.1941.183.110 |s2cid=118624331 }}{{Cite journal |title=The Lemniscate Constants |last=Todd |first=John |journal=Communications of the ACM |volume=18 |number=1 |year=1975 |pages=14–19 |doi=10.1145/360569.360580 |s2cid=85873 |doi-access=free }} The set $\backslash \{\backslash pi,M(1,1/\backslash sqrt\{2\})\backslash \}$ is algebraically independent over $\backslash mathbb\{Q\}$,G. V. Choodnovsky: Algebraic independence of constants connected with the functions of analysis, Notices of the AMS 22, 1975, p. A-486G. V. Chudnovsky: Contributions to The Theory of Transcendental Numbers, American Mathematical Society, 1984, p. 6 but the set $\backslash \{\backslash pi,M(1,1/\backslash sqrt\{2\}),M\text{'}(1,1/\backslash sqrt\{2\})\backslash \}$ (where the prime denotes the derivative with respect to the second variable) is not algebraically independent over $\backslash mathbb\{Q\}$. In fact,{{Cite book |last1=Borwein |first1=Jonathan M. |last2=Borwein| first2=Peter B. |title=Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity |publisher=Wiley-Interscience |year=1987 |edition=First |isbn=0-471-83138-7}} p. 45$$\backslash pi=2\backslash sqrt\{2\}\backslash frac\{M^3(1,1/\backslash sqrt\{2\})\}\{M\text{'}(1,1/\backslash sqrt\{2\})\}.$$The geometric–harmonic mean GH can be calculated using analogous sequences of geometric and harmonic means, and in fact {{math|1= GH(x, y) = 1/M(1/x, 1/y) = xy/M(x, y)}}.{{cite journal

| last = Newman | first = D. J.

| doi = 10.2307/2007804

| journal = Mathematics of Computation

| pages = 207–210

| title = A simplified version of the fast algorithms of Brent and Salamin

| volume = 44

| year = 1985| issue = 169

| jstor = 2007804

}}

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,{{AS ref|17|598–599}} and Jacobi elliptic functions.{{cite book |first=Louis V. |last=King |author-link=Louis Vessot King |url=https://archive.org/details/onthenumerical032686mbp |title=On the Direct Numerical Calculation of Elliptic Functions and Integrals |publisher=Cambridge University Press |year=1924 }}

The inequality of arithmetic and geometric means implies that$$g\_n\; \backslash leq\; a\_n$$and thus$$g\_\{n\; +\; 1\}\; =\; \backslash sqrt\{g\_n\; \backslash cdot\; a\_n\}\; \backslash geq\; \backslash sqrt\{g\_n\; \backslash cdot\; g\_n\}\; =\; g\_n$$that is, the sequence {{math|g_n}} is nondecreasing and bounded above by the larger of {{math|x}} and {{math|y}}. By the monotone convergence theorem, the sequence is convergent, so there exists a {{math|g}} such that:$$\backslash lim\_\{n\backslash to\; \backslash infty\}g\_n\; =\; g$$However, we can also see that:$$a\_n\; =\; \backslash frac\{g\_\{n\; +\; 1\}^2\}\{g\_n\}$$

and so:

$$\backslash lim\_\{n\backslash to\; \backslash infty\}a\_n\; =\; \backslash lim\_\{n\backslash to\; \backslash infty\}\backslash frac\{g\_\{n\; +\; 1\}^2\}\{g\_\{n\}\}\; =\; \backslash frac\{g^2\}\{g\}\; =\; g$$

Q.E.D.

This proof is given by Gauss.

Let

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

Changing the variable of integration to $\backslash theta\text{'}$, where

$$\backslash sin\backslash theta\; =\; \backslash frac\{2x\backslash sin\backslash theta\text{'}\}\{(x+y)+(x-y)\backslash sin^2\backslash theta\text{'}\}\; \backslash Rightarrow\; d(\backslash sin\backslash theta)=d\backslash left(\backslash frac\{2x\backslash sin\backslash theta\text{'}\}\{(x+y)+(x-y)\backslash sin^2\backslash theta\text{'}\}\backslash right)\backslash Rightarrow\; \backslash cos\backslash theta\backslash \; d\backslash theta\; =2x\; \backslash frac\{(x+y)-(x-y)\backslash sin^2\backslash theta\text{'}\}\{((x+y)+(x-y)\backslash sin^2\backslash theta\text{'})^2\}\backslash \; \backslash cos\backslash theta\text{'}\; d\backslash theta\text{'}$$

$$\backslash cos\backslash theta\; =\; \backslash frac\{\backslash sqrt\{(x+y)^2-2(x^2+y^2)\backslash sin^2\backslash theta\text{'}+(x-y)^2\backslash sin^4\backslash theta\text{'}\}\}\{(x+y)+(x-y)\backslash sin^2\backslash theta\text{'}\}=\backslash frac\{\backslash cos\backslash theta\text{'}\backslash sqrt\{(x-y)^2\backslash cos^2\backslash theta\text{'}+4xy\}\}\{(x+y)+(x-y)\backslash sin^2\backslash theta\text{'}\}=\backslash frac\{\backslash cos\backslash theta\text{'}\backslash sqrt\{(x+y)^2\backslash cos^2\backslash theta\text{'}+4xy\backslash sin^2\backslash theta\text{'}\}\}\{(x+y)+(x-y)\backslash sin^2\backslash theta\text{'}\}\; ,$$

$$\backslash Rightarrow\; \backslash cos\backslash theta\backslash \; d\backslash theta\; =\backslash frac\{\backslash cos\backslash theta\text{'}\backslash sqrt\{(x+y)^2\backslash cos^2\backslash theta\text{'}+4xy\backslash sin^2\backslash theta\text{'}\}\}\{(x+y)+(x-y)\backslash sin^2\backslash theta\text{'}\}\backslash \; d\backslash theta\; =2x\; \backslash frac\{(x+y)-(x-y)\backslash sin^2\backslash theta\text{'}\}\{((x+y)+(x-y)\backslash sin^2\backslash theta\text{'})^2\}\backslash \; \backslash cos\backslash theta\text{'}\; d\backslash theta\text{'}\; ,$$

$$\backslash Rightarrow\; d\backslash theta\; =\; \backslash frac\{x((x+y)-(x-y)\backslash sin^2\backslash theta\text{'})\}\{((x+y)+(x-y)\backslash sin^2\backslash theta\text{'})\}\; \backslash frac\{2\; d\backslash theta\text{'}\}\{\backslash sqrt\{(x+y)^2\backslash cos^2\backslash theta\text{'}+4xy\backslash sin^2\backslash theta\text{'}\}\}$$

\ ,

$$\backslash sqrt\{x^2\backslash cos^2\backslash theta+y^2\backslash sin^2\backslash theta\}\; =\; \backslash frac\{\backslash sqrt\{\{x^2\; ((x+y)^2-2(x^2+y^2)\backslash sin^2\backslash theta\text{'}+(x-y)^2\backslash sin^4\backslash theta\text{'})+4x^2y^2\backslash sin^2\backslash theta\text{'}\}\}\}\{((x+y)+(x-y)\backslash sin^2\backslash theta\text{'})\}=\; \backslash frac\{x\; ((x+y)-(x-y)\backslash sin^2\backslash theta\text{'})\}\{((x+y)+(x-y)\backslash sin^2\backslash theta\text{'})\}$$

This yields

$$\backslash frac\{d\backslash theta\}\{\backslash sqrt\{x^2\backslash cos^2\backslash theta+y^2\backslash sin^2\backslash theta\}\}\; =\; \backslash frac\{2\; d\backslash theta\text{'}\}\{\backslash sqrt\{(x+y)^2\backslash cos^2\backslash theta\text{'}+4xy\backslash sin^2\backslash theta\text{'}\}\}\; =\; \backslash frac\{d\backslash theta\text{'}\}\{\backslash sqrt\{((\backslash frac\{x+y\}\{2\})^2\backslash cos^2\backslash theta\text{'}+(\backslash sqrt\{xy\})^2\backslash sin^2\backslash theta\text{'}\}\}$$

,

gives

$$\backslash begin\{align\}$$

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 $I(z,z)\; =\; \backslash pi/(2z)$.

Finally, we obtain the desired result

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

According to the Gauss–Legendre algorithm,{{cite journal |first=Eugene |last=Salamin |author-link=Eugene Salamin (mathematician) |title=Computation of π using arithmetic–geometric mean |journal=Mathematics of Computation |url=https://link.springer.com/chapter/10.1007/978-3-319-32377-0_1 |volume=30 |issue=135 |year=1976 |pages=565–570 |doi=10.2307/2005327 |jstor=2005327 |mr=0404124 }}

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

where

$$c\_j\; =\; \backslash frac\{1\}\{2\}\backslash left(a\_\{j-1\}-g\_\{j-1\}\backslash right)\; ,$$

with $a\_0=1$ and $g\_0=1/\backslash sqrt\{2\}$, which can be computed without loss of precision using

$$c\_j\; =\; \backslash frac\{c\_\{j-1\}^2\}\{4a\_j\}\; .$$

Taking $a\_0\; =\; 1$ and $g\_0\; =\; \backslash cos\backslash alpha$ yields the AGM

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

where {{math|K(k)}} is a complete elliptic integral of the first kind:

$$K(k)\; =\; \backslash int\_0^\{\backslash pi/2\}(1\; -\; k^2\; \backslash sin^2\backslash theta)^\{-1/2\}\; \backslash ,\; d\backslash theta.$$

That is to say that this quarter period may be efficiently computed through the AGM,

$$K(k)\; =\; \backslash frac\{\backslash pi\}\{2M(1,\backslash sqrt\{1-k^2\})\}\; .$$

Using this property of the AGM along with the ascending transformations of John Landen,{{cite journal |first=John |last=Landen |title=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 |journal=Philosophical Transactions of the Royal Society |volume=65 |year=1775 |pages=283–289 |doi=10.1098/rstl.1775.0028|s2cid=186208828 }} Richard P. Brent{{cite journal |first=Richard P. |last=Brent |title=Fast Multiple-Precision Evaluation of Elementary Functions |journal=Journal of the ACM |volume=23 |issue=2 |year=1976 |pages=242–251 |doi=10.1145/321941.321944 |mr=0395314 |citeseerx=10.1.1.98.4721 |s2cid=6761843 |url=https://link.springer.com/chapter/10.1007/978-3-319-32377-0_2 }} suggested the first AGM algorithms for the fast evaluation of elementary transcendental functions ({{math|e^x}}, {{math|cos x}}, {{math|sin x}}). Subsequently, many authors went on to study the use of the AGM algorithms.{{cite book |author1-link=Jonathan Borwein |first1=Jonathan M. |last1=Borwein |author2-link=Peter Borwein |first2=Peter B. |last2=Borwein |title=Pi and the AGM |publisher=Wiley |place=New York |year=1987 |isbn=0-471-83138-7 |mr=0877728 }}

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

{{reflist|group=note}}

{{reflist}}

{{refbegin}}

• {{cite journal |author1-last=Daróczy |author1-first=Zoltán |author2-last=Páles |author2-first=Zsolt |year=2002 |title=Gauss-composition of means and the solution of the Matkowski–Suto problem |journal=Publicationes Mathematicae Debrecen |volume=61 |issue=1–2 |pages=157–218 |doi=10.5486/PMD.2002.2713 }}
• {{SpringerEOM |title=Arithmetic–geometric mean process|mode=cs1}}
• {{mathworld |urlname=Arithmetic-GeometricMean}}

{{refend}}

{{Statistics|descriptive}}

{{DEFAULTSORT:Arithmetic-Geometric Mean}}

Category:Means

Category:Special functions

Category:Elliptic functions

Category:Articles containing proofs