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

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

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

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

Then if we define

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

we have

$$F\left(2\right)=\frac{\pi}{2}\\$$

but if we search for more relations we also find that

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

where $$\kappa=1.6999683849605364588$$… this gives the identity

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

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,

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

and

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

and as

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

we have the closed form

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

but by letting

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

It would appear that

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

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