How do I derive the relationship between the Arithmetic-Geometric Mean and an elliptic integral?

Question

kainui · Answer

$$\Large K(z) = \int\limits_0^{\pi/2} \frac{d \theta}{ \sqrt{1-z^2\sin^2 \theta}}$$ Apparently we can rewrite this integral as $$\Large K(z) = \frac{\pi}{2} \frac{1}{AGM(z+1, 1-z)}$$ where AGM is the arithemetic-geometric mean which is the limit of this process: $$\Large A_{n+1} = \frac{A_n+B_n}{2} \ \Large B_{n+1} = \sqrt{A_n B_n}$$

ganeshie8 · Answer

$$K(z) = I(1, z) = \dfrac{\pi}{2AGM(1, \sqrt{1-z^2})} = \dfrac{\pi}{2AGM(1+z, 1-z)}$$ you got that AGM(1+z, 1-z) thingy like this ? this looks like a corollary to Gauss's theorem http://en.wikipedia.org/wiki/Arithmetic%E2%80%93geometric_mean