# §19.8(i) Gauss’s Arithmetic-Geometric Mean (AGM)

When $a_{0}$ and $g_{0}$ are positive numbers, define

 19.8.1 $\displaystyle a_{n+1}$ $\displaystyle=\frac{a_{n}+g_{n}}{2},$ $\displaystyle g_{n+1}$ $\displaystyle=\sqrt{a_{n}g_{n}}$, $n=0,1,2,\dots$. Symbols: $n$: nonnegative integer, $a_{n}$: iterate and $g_{n}$: iterate Permalink: http://dlmf.nist.gov/19.8.E1 Encodings: TeX, TeX, pMML, pMML, png, png

As $n\to\infty$, $a_{n}$ and $g_{n}$ converge to a common limit $\mathop{M\/}\nolimits\!\left(a_{0},g_{0}\right)$ called the AGM (Arithmetic-Geometric Mean) of $a_{0}$ and $g_{0}$. By symmetry in $a_{0}$ and $g_{0}$ we may assume $a_{0}\geq g_{0}$ and define

 19.8.2 $c_{n}=\sqrt{a_{n}^{2}-g_{n}^{2}}.$

Then

 19.8.3 $c_{n+1}=\frac{a_{n}-g_{n}}{2}=\frac{c_{n}^{2}}{4a_{n+1}},$

showing that the convergence of $c_{n}$ to 0 and of $a_{n}$ and $g_{n}$ to $\mathop{M\/}\nolimits\!\left(a_{0},g_{0}\right)$ is quadratic in each case.

The AGM has the integral representations

 19.8.4 $\frac{1}{\mathop{M\/}\nolimits\!\left(a_{0},g_{0}\right)}=\frac{2}{\pi}\int_{0% }^{\pi/2}\frac{d\theta}{\sqrt{a_{0}^{2}{\mathop{\cos\/}\nolimits^{2}}\theta+g_% {0}^{2}{\mathop{\sin\/}\nolimits^{2}}\theta}}=\frac{1}{\pi}\int_{0}^{\infty}% \frac{dt}{\sqrt{t(t+a_{0}^{2})(t+g_{0}^{2})}}.$

The first of these shows that

 19.8.5 $\mathop{K\/}\nolimits\!\left(k\right)=\frac{\pi}{2\!\mathop{M\/}\nolimits\!% \left(1,k^{\prime}\right)},$ $-\infty.

The AGM appears in

 19.8.6 $\mathop{E\/}\nolimits\!\left(k\right)=\frac{\pi}{2\!\mathop{M\/}\nolimits\!% \left(1,k^{\prime}\right)}\left(a_{0}^{2}-\sum_{n=0}^{\infty}2^{n-1}c_{n}^{2}% \right)=\mathop{K\/}\nolimits\!\left(k\right)\left(a_{1}^{2}-\sum_{n=2}^{% \infty}2^{n-1}c_{n}^{2}\right),$ $-\infty, $a_{0}=1$, $g_{0}=k^{\prime}$,

and in

 19.8.7 $\mathop{\Pi\/}\nolimits\!\left(\alpha^{2},k\right)=\frac{\pi}{4\!\mathop{M\/}% \nolimits\!\left(1,k^{\prime}\right)}\left(2+\frac{\alpha^{2}}{1-\alpha^{2}}% \sum_{n=0}^{\infty}Q_{n}\right),$ $-\infty, $-\infty<\alpha^{2}<1$,

where $a_{0}=1$, $g_{0}=k^{\prime}$, $p_{0}^{2}=1-\alpha^{2}$, $Q_{0}=1$, and

 19.8.8 $\displaystyle p_{n+1}$ $\displaystyle=\frac{p_{n}^{2}+a_{n}g_{n}}{2p_{n}},$ $\displaystyle\varepsilon_{n}$ $\displaystyle=\frac{p_{n}^{2}-a_{n}g_{n}}{p_{n}^{2}+a_{n}g_{n}},$ $\displaystyle Q_{n+1}$ $\displaystyle=\tfrac{1}{2}Q_{n}\varepsilon_{n}$, $n=0,1,\dots$.

Again, $p_{n}$ and $\varepsilon_{n}$ converge quadratically to $\mathop{M\/}\nolimits\!\left(a_{0},g_{0}\right)$ and 0, respectively, and $Q_{n}$ converges to 0 faster than quadratically. If $\alpha^{2}>1$, then the Cauchy principal value is

 19.8.9 $\mathop{\Pi\/}\nolimits\!\left(\alpha^{2},k\right)=\frac{\pi}{4\!\mathop{M\/}% \nolimits\!\left(1,k^{\prime}\right)}\frac{k^{2}}{k^{2}-\alpha^{2}}\sum_{n=0}^% {\infty}Q_{n},$ $-\infty, $1<\alpha^{2}<\infty$,

where (19.8.8) still applies, but with

 19.8.10 $p_{0}^{2}=1-(k^{2}/\alpha^{2}).$ Symbols: $k$: real or complex modulus and $\alpha^{2}$: real or complex parameter Permalink: http://dlmf.nist.gov/19.8.E10 Encodings: TeX, pMML, png

# Descending Landen Transformation

Let

 19.8.11 $\displaystyle k_{1}$ $\displaystyle=\frac{1-k^{\prime}}{1+k^{\prime}},$ $\displaystyle\phi_{1}$ $\displaystyle=\phi+\mathop{\mathrm{arctan}\/}\nolimits\!\left(k^{\prime}% \mathop{\tan\/}\nolimits\phi\right)$ $\displaystyle=\mathop{\mathrm{arcsin}\/}\nolimits\!\left((1+k^{\prime})\frac{% \mathop{\sin\/}\nolimits\phi\mathop{\cos\/}\nolimits\phi}{\sqrt{1-k^{2}{% \mathop{\sin\/}\nolimits^{2}}\phi}}\right).$

(Note that $0 and $0<\phi<\pi/2$ imply $k_{1} and $\phi<\phi_{1}<2\phi$, and also that $\phi=\pi/2$ implies $\phi_{1}=\pi$.) Then

 19.8.12 $\displaystyle\mathop{K\/}\nolimits\!\left(k\right)$ $\displaystyle=(1+k_{1})\mathop{K\/}\nolimits\!\left(k_{1}\right),$ $\displaystyle\mathop{E\/}\nolimits\!\left(k\right)$ $\displaystyle=(1+k^{\prime})\mathop{E\/}\nolimits\!\left(k_{1}\right)-k^{% \prime}\mathop{K\/}\nolimits\!\left(k\right).$
 19.8.13 $\displaystyle\mathop{F\/}\nolimits\!\left(\phi,k\right)$ $\displaystyle=\tfrac{1}{2}(1+k_{1})\mathop{F\/}\nolimits\!\left(\phi_{1},k_{1}% \right),$ $\displaystyle\mathop{E\/}\nolimits\!\left(\phi,k\right)$ $\displaystyle=\tfrac{1}{2}(1+k^{\prime})\mathop{E\/}\nolimits\!\left(\phi_{1},% k_{1}\right)-k^{\prime}\mathop{F\/}\nolimits\!\left(\phi,k\right)+\tfrac{1}{2}% (1-k^{\prime})\mathop{\sin\/}\nolimits\phi_{1}.$
 19.8.14 $2(k^{2}-\alpha^{2})\mathop{\Pi\/}\nolimits\!\left(\phi,\alpha^{2},k\right)=% \frac{\omega^{2}-\alpha^{2}}{1+k^{\prime}}\mathop{\Pi\/}\nolimits\!\left(\phi_% {1},\alpha_{1}^{2},k_{1}\right)+k^{2}\mathop{F\/}\nolimits\!\left(\phi,k\right% )-{(1+k^{\prime})\alpha_{1}^{2}\mathop{R_{C}\/}\nolimits\!\left(c_{1},c_{1}-% \alpha_{1}^{2}\right)},$

where

 19.8.15 $\displaystyle\omega^{2}$ $\displaystyle=\frac{k^{2}-\alpha^{2}}{1-\alpha^{2}},$ $\displaystyle\alpha_{1}^{2}$ $\displaystyle=\frac{\alpha^{2}\omega^{2}}{(1+k^{\prime})^{2}},$ $\displaystyle c_{1}$ $\displaystyle={\mathop{\csc\/}\nolimits^{2}}\phi_{1}.$

# Ascending Landen Transformation

Let

 19.8.16 $\displaystyle k_{2}$ $\displaystyle=2\sqrt{k}/(1+k),$ $\displaystyle 2\phi_{2}$ $\displaystyle=\phi+\mathop{\mathrm{arcsin}\/}\nolimits\!\left(k\mathop{\sin\/}% \nolimits\phi\right).$

(Note that $0 and $0<\phi\leq\pi/2$ imply $k and $\phi_{2}<\phi$.) Then

 19.8.17 $\displaystyle\mathop{F\/}\nolimits\!\left(\phi,k\right)$ $\displaystyle=\frac{2}{1+k}\mathop{F\/}\nolimits\!\left(\phi_{2},k_{2}\right),$ $\displaystyle\mathop{E\/}\nolimits\!\left(\phi,k\right)$ $\displaystyle=(1+k)\mathop{E\/}\nolimits\!\left(\phi_{2},k_{2}\right)+(1-k)% \mathop{F\/}\nolimits\!\left(\phi_{2},k_{2}\right)-k\mathop{\sin\/}\nolimits\phi.$

# §19.8(iii) Gauss Transformation

We consider only the descending Gauss transformation because its (ascending) inverse moves $\mathop{F\/}\nolimits\!\left(\phi,k\right)$ closer to the singularity at $k=\mathop{\sin\/}\nolimits\phi=1$. Let

 19.8.18 $\displaystyle k_{1}$ $\displaystyle=(1-k^{\prime})/(1+k^{\prime}),$ $\displaystyle\mathop{\sin\/}\nolimits\psi_{1}$ $\displaystyle=\frac{(1+k^{\prime})\mathop{\sin\/}\nolimits\phi}{1+\Delta},$ $\displaystyle\Delta$ $\displaystyle=\sqrt{1-k^{2}{\mathop{\sin\/}\nolimits^{2}}\phi}.$

(Note that $0 and $0<\phi<\pi/2$ imply $k_{1} and $\psi_{1}<\phi$, and also that $\phi=\pi/2$ implies $\psi_{1}=\pi/2$, thus preserving completeness.) Then

 19.8.19 $\displaystyle\mathop{F\/}\nolimits\!\left(\phi,k\right)$ $\displaystyle=(1+k_{1})\mathop{F\/}\nolimits\!\left(\psi_{1},k_{1}\right),$ $\displaystyle\mathop{E\/}\nolimits\!\left(\phi,k\right)$ $\displaystyle=(1+k^{\prime})\mathop{E\/}\nolimits\!\left(\psi_{1},k_{1}\right)% -k^{\prime}\mathop{F\/}\nolimits\!\left(\phi,k\right)+(1-\Delta)\mathop{\cot\/% }\nolimits\phi,$
 19.8.20 $\rho\mathop{\Pi\/}\nolimits\!\left(\phi,\alpha^{2},k\right)=\frac{4}{1+k^{% \prime}}\mathop{\Pi\/}\nolimits\!\left(\psi_{1},\alpha^{2}_{1},k_{1}\right)+(% \rho-1)\mathop{F\/}\nolimits\!\left(\phi,k\right)-\mathop{R_{C}\/}\nolimits\!% \left(c-1,c-\alpha^{2}\right),$

where

 19.8.21 $\displaystyle\rho$ $\displaystyle=\sqrt{1-(k^{2}/\alpha^{2})},$ $\displaystyle\alpha_{1}^{2}$ $\displaystyle=\alpha^{2}(1+\rho)^{2}/(1+k^{\prime})^{2},$ $\displaystyle c$ $\displaystyle={\mathop{\csc\/}\nolimits^{2}}\phi.$

If $0<\alpha^{2}, then $\rho$ is pure imaginary.