# Connection between Infinite continued fractions, elliptic integrals and AGM

It is known that at $x=1$, the following continued fraction represents $\frac{4}{\pi}$ and can be approximated rapidly using Gauss' Arithmetic Geometric mean. $$C(x) = x + \frac{1^{2}}{2x + \frac{3^{2}}{2x + \frac{5^{2}}{2x + \frac{7^{2}}{2x + \cdots}}}}$$

Are there any other $x$ that $C(x)$ can be approximated through AGM quickly? Is there any connection to elliptic integrals?

• Mathworld says that $C(1) = 4/\pi$, not $\pi$. – S. Carnahan Dec 23 '12 at 5:30
• I meant represents in a crude way. I can correct it. – Turbo Dec 23 '12 at 8:30
• Okay, sorry to jump on you like that. It is a good question. – S. Carnahan Dec 23 '12 at 15:35
• The AGM doesn't make any use of that continued fraction for $4/\pi$, does it? So there's no reason to think there's any connection between AGM and $C(x)$, right? – Gerry Myerson Jun 26 '13 at 1:58
• $C(4n-3),~n\in\mathbb{N}$ is a rational multiple of $1/\pi$. – Nemo Feb 23 '17 at 18:38

As was noted in the comments, the function $C(x)$ can be expressed in terms of the Gamma function, see Corollary 1 on page 145 in Ramanujan's Notebook II which attributes this result to G. Bauer: Von einem Kettenbruche Euler's und einem Theorem von Wallis, Abh. Bayer. Akad. Wiss. 11(1872), 96 - 116 and to the first Ramanujan letter to Hardy.

$$C(x) = \frac{\Gamma^2(\frac{x+1}{4})}{\Gamma^2(\frac{x+3}{4})}$$

The AGM function can be expressed in terms of complete elliptic integral of the first kind, as is noted on the Wikipedia page, which is given by Gauss hypergeometric function ${}_2F_1(-\frac{1}{2}, \frac{1}{2}; 1; k^2)$ whereas the function $C(x)$ is given by ${}_2F_1(-\frac{1}{2}, -\frac{1}{2};x,1)$.

There is certainly more to the story, see related question on math.stackexchange and equation (99) at MathWorld.

The explicit relation connecting Gamma values at certain rational points and elliptic integral can be found e.g. in the article of Borwein and Zucker, book Pi and the AGM by Borwein, Borwein and a more recent article by Vidunas.

• Note that the reference you cite attributes this corollary to Bauer 1872. – Noam D. Elkies Feb 27 '17 at 0:53
• @VitTucek thank you. do you know how fast $(99)$ converges? is it as fast as AGM? – Turbo Feb 27 '17 at 7:03
• @Turbo My understanding is that it also converges quadratically. – Vít Tuček Feb 27 '17 at 10:16