Arithmetic Geometric Mean Formulae

Abstract

I briefly show how a whole set of formula mimicking the Gauss-Salamin formula can be derived associated with singular values of the complete elliptic integral \(K(k^{2})\), (with elliptic modulus \(k\)).

Main

From the Gauss–Salamin formula for \(\pi\), it can be easily shown that

\begin{equation} \frac{\pi}{2}=\mathrm{agm}\left(1,\frac{1}{\sqrt{2}}\right)K\left(\frac{1}{\sqrt{2}}\right)\\ \end{equation}

where \(K\) is a complete elliptic integral of the first kind with modulus \(k=\frac{1}{\sqrt{2}}\) and \(\mathrm{agm}(x,y)\) is the arithmetic geometric mean function, which is the shared limit of \(a_{\infty}\) or \(g_{\infty}\) of the iterative formula

\begin{equation} a_{0}=\frac{1}{2}(x+y)\\ g_{0}=\sqrt{xy}\\ a_{n}=\frac{1}{2}(a_{n-1}+g_{n-1})\\ g_{n}=\sqrt{a_{n-1}g_{n-1}}\\ \end{equation}

Then if we define

\begin{equation} F(x)=\mathrm{agm}\left(1,\frac{1}{\sqrt{x}}\right)K\left(\frac{1}{\sqrt{x}}\right)=\frac{\pi}{4}\frac{1+\sqrt{x}}{\sqrt{x}}\frac{K\left(\frac{1}{\sqrt{x}}\right)}{K\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)}\\ \end{equation}

we have

\begin{equation} F\left(2\right)=\frac{\pi}{2}\\ \end{equation}

but if we search for more relations we also find that

\begin{equation} F\left(\frac{1}{8}\left(4+3\sqrt{2}\right)\right)=\pi\\ F\left(17+12\sqrt{2}\right)=F\left(\left(1+\sqrt{2}\right)^{4}\right)=\frac{\pi}{4}\\ F\left(\left(1+\sqrt{2}\right)^{2}\right)=\frac{\pi}{2\sqrt{2}}\\ F\left(2-\sqrt{2}\right)=\kappa-\frac{\pi}{2}i\\ F\left(1+\frac{1}{\sqrt{2}}\right)=\kappa\\ \end{equation}

where \(\kappa=1.6999683849605364588\)… this gives the identity

\begin{equation} F\left(2-\sqrt{2}\right)=F\left(1+\frac{1}{\sqrt{2}}\right)-iF(2)\\ \end{equation}

we see that these identities are in fact independent of \(\pi\) as the values cancel from equation \(3\). We discover from these various identities about the complete elliptic integral of the first kind,

\begin{equation} \frac{K\left(\frac{\sqrt{8}}{\sqrt{4+3\sqrt{2}}}\right)}{K\left(\frac{\sqrt{8+6\sqrt{2}}-4}{\sqrt{8+6\sqrt{2}}+4}\right)}=\frac{\sqrt{8+6\sqrt{2}}}{1+\frac{1}{2}\sqrt{2+\frac{3}{2}\sqrt{2}}}\\ \end{equation}

and

\begin{equation} \frac{K\left(3-2\sqrt{2}\right)}{K\left(\frac{1}{\sqrt{2}}\right)}=\frac{1}{4}(2+\sqrt{2})\\ \end{equation}

and as

\begin{equation} K\left(\frac{1}{\sqrt{2}}\right)=\frac{8\pi^{3/2}}{\Gamma(-\frac{1}{4})^{2}}\\ \end{equation}

we have the closed form

\begin{equation} K\left(3-2\sqrt{2}\right)=\frac{2(2+\sqrt{2})\pi^{3/2}}{\Gamma\left(-\frac{1}{4}\right)^{2}}=1.58255...\\ \end{equation}

but by letting

\begin{equation} a=\frac{4+3\sqrt{2}}{8}\\ \end{equation} \begin{equation} \frac{K\left(\frac{1}{\sqrt{a}}\right)}{K\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)}=\frac{4\sqrt{a}}{1+\sqrt{a}}\\ \end{equation}

It would appear that

\begin{equation} \mathrm{agm}(\frac{1}{2},\frac{1}{\sqrt{2}})=\frac{\Gamma(3/4)}{\Gamma(1/4)}\sqrt{\pi}\\ \end{equation}

this directly leads to

\begin{equation} K(2\sqrt{2}-3)=\frac{(1+\sqrt{2})\sqrt{\pi}\Gamma(1/4)}{8\Gamma(3/4)}\\ \end{equation}

Why is this working

We see a correspondence between special points from http://mathworld.wolfram.com/EllipticIntegralSingularValue.html.. Using the identities there we can then write

\begin{equation} \mathrm{agm}(1,3+2\sqrt{2})=\frac{\pi^{3/2}(4+2\sqrt{2})}{\Gamma(\frac{1}{4})^{2}}\\ \mathrm{agm}(1,1+\sqrt{2})=\frac{2\sqrt{4+2\sqrt{2}}\pi^{3/2}}{\Gamma(\frac{1}{8})\Gamma(\frac{3}{8})}\\ \mathrm{agm}(1,\frac{8}{4+\sqrt{2}-\sqrt{6}}-1)=\frac{8\cdot 2^{1/3}\pi^{2}}{3^{1/4}(4+\sqrt{2}-\sqrt{6})\Gamma(\frac{1}{3})^{3}}\\ \end{equation}

but actually we may write a more general form for each of these, for example

\begin{equation} \mathrm{agm}(y,(15-10\sqrt{2}-8\sqrt{3}+6\sqrt{6})y)=\frac{2^{4/3}(8-5\sqrt{2}-4\sqrt{3}+3\sqrt{6})\pi^{2}y}{3^{1/4}\Gamma(\frac{1}{3})^{3}},\forall y\in\mathbb{R}\\ \end{equation}

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