π(PI) and the AGM: Evaluating Elliptic Integrals

Complete Elliptic Integrals of First Kind

In my earlier post I described the method for calculating complete elliptic integrals of first kind namely $ K(k)$ and $ I(a, b)$. To summarize the results we have
$$ K(k) = \int_{0}^{\pi / 2}\frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}} = \frac{\pi}{2M(1, \sqrt{1 - k^{2}})} = \frac{\pi}{2M(1, k^{\prime})}$$ $$ I(a, b) = \int_{0}^{\pi / 2}\frac{d\theta}{\sqrt{a^{2}\cos^{2}\theta + b^{2}\sin^{2}\theta}} = \frac{\pi}{2M(a, b)}$$ where $ M(a, b)$ denotes the Arithmetic-Geometric Mean of two numbers $ a$ and $ b$ and $ k^{\prime}$ is the complementary modulus related to $ k$ by the following relation $$ k^{2} + k^{\prime 2} = 1$$ Here we shall discuss the problem of evaluating the complete elliptic integrals of second kind, i.e. $ E(k)$ or $ J(a, b)$. This is a hard problem compared to that of evaluating $ K(k)$ and involves a somewhat complicated formula although it is also based on the ever important concept of AGM.

Arithmetic-Geometric Mean Sequences

To begin with we have to focus not only on the AGM concept but also the associated AGM sequences. So let's assume two positive numbers $ a$ and $ b$ with $ a > b$ and form the sequences $ \{a_{n}\}, \{b_{n}\}, \{c_{n}\}$ as follows $$ a_{0} = a,\,\, b_{0} = b,\,\, c_{1} = \frac{a - b}{2}$$ $$ a_{n+1} = \frac{a_{n} + b_{n}}{2},\,\, b_{n + 1} = \sqrt{a_{n}b_{n}},\,\, c_{n + 1} = \frac{a_{n} - b_{n}}{2}$$ Also note that the sequence $ \{c_{n}\}$ starts with $ n = 1$ (this can be remedied by defining the sequences in reverse direction with negative indexes, but we can do without that stuff). Then it has been already established (see the post linked above) that sequences $ \{a_{n}\}$ and $ \{b_{n}\}$ tend to a common limit $ M(a, b)$ and the sequence $ \{c_{n}\}$ therefore tends to $ 0$.

The sequence $ \{c_{n}\}$ is important in evaluating the elliptic integrals of second kind and hence it is better to list its common properties

1. $ a_{n} = a_{n + 1} + c_{n + 1}$
2. $ b_{n} = a_{n + 1} - c_{n + 1}$
3. $ c_{n}^{2} = a_{n}^{2} - b_{n}^{2}\,\,$ (this defines $ c_{0}$)
4. $ a_{n}c_{n} = \dfrac{c_{n - 1}^{2}}{4}\,\,$ (this can define $ c_{n}$ for negative $ n$).

The above properties can be established easily from the given definitions. Armed with the basics of AGM sequences we can move on to consider the Landen Transformation which is the key in evaluation of elliptic integrals.

Landen Transformation

Landen Transformation is a susbtitution formula used to effect a change of variables in the elliptic integrals to transform the integral into simpler and more amenable form. These transformations were discovered by a somewhat obscure mathematician John Landen in 18th century. Landen transformation is central to the numerical evaluation of elliptic integrals and is intimately related to the AGM, however Landen did not forsee them and it was left for Legendre and Gauss to build up on these transformations and devise equivalent transformations needed to evaluate elliptic integrals.

It also needs to be noted that none of these transformations is simple in terms of the algebraic manipulations involved, but the end result is very simple, beautiful and striking. The following Landen's transformation $$ \sin(2\phi - \theta) = k\sin\theta\tag{1}$$ leads to the beautiful result $$ K(k) = \frac{1}{1 + k}K\left(\frac{2\sqrt{k}}{1 + k}\right)\tag{2}$$ which is equivalent to the simple property of AGM given by \begin{align}M(1 + x, 1 - x) &= (1 + x)M\left(1, \frac{1 - x}{1 + x}\right)\notag\\ &= (1 + x)M\left(1 + \frac{2\sqrt{x}}{1 + x}, 1 - \frac{2\sqrt{x}}{1 + x}\right)\notag\end{align} To prove that this is so we start with the Landen transformation to get $$ \sin(2\phi) \cos\theta - \cos(2\phi) \sin\theta = k\sin\theta $$ $$ \Rightarrow \tan\theta = \frac{\sin(2\phi)}{k + \cos(2\phi)}\tag{3}$$ and then we have \begin{align}\cos\theta &= \frac{1}{\sqrt{1 + \tan^{2}\theta}}\notag\\ &= \frac{k + \cos(2\phi)}{\sqrt{(k + \cos(2\phi))^{2} + \sin^{2}(2\phi)}}\notag\\ &= \frac{k + \cos(2\phi)}{\sqrt{1 + k^{2} + 2k\cos(2\phi)}}\notag\end{align} Similarly $$ \sin\theta = \tan\theta \cdot \cos\theta = \frac{\sin(2\phi)}{\sqrt{1 + k^{2} + 2k\cos(2\phi)}}$$ and therefore \begin{align}\cos(2\phi - \theta) &= \cos(2\phi) \cos\theta + \sin(2\phi) \sin\theta\notag\\ &= \frac{\cos(2\phi) (k + \cos(2\phi)) + \sin^{2}(2\phi)}{\sqrt{1 + k^{2} + 2k\cos(2\phi)}}\notag\\ &= \frac{1 + k\cos(2\phi)}{\sqrt{1 + k^{2} + 2k\cos(2\phi)}}\notag\end{align} Finally differentiating $(3)$ we get \begin{align}\frac{d\theta}{d\phi} &= \cos^{2}\theta \cdot \frac{2(k + \cos(2\phi))\cos(2\phi) + 2\sin^{2}(2\phi)}{(k + \cos(2\phi))^{2}}\notag\\ &= \frac{2(1 + k\cos(2\phi))}{1 + k^{2} + 2k\cos(2\phi)}\notag\end{align} Now with all the substitution formulas in place we have \begin{align}K(k) &= \int_{0}^{\pi / 2}\frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}} = \frac{1}{2}\int_{0}^{\pi}\frac{d\theta}{\sqrt{1 - k^{2}\sin^{2}\theta}}\notag\\ &= \frac{1}{2}\int_{0}^{\pi / 2}\frac{\sqrt{1 + k^{2} + 2k\cos(2\phi)}}{1 + k\cos(2\phi)}\frac{2(1 + k\cos(2\phi))}{1 + k^{2} + 2k\cos(2\phi)}\,d\phi\notag\\ &= \int_{0}^{\pi / 2}\frac{d\phi}{\sqrt{1 + k^{2} + 2k - 4k\sin^{2}\phi}}\notag\\ &= \int_{0}^{\pi / 2}\frac{d\phi}{\sqrt{(1 + k)^{2} - 4k\sin^{2}\phi}}\notag\\ &= \frac{1}{1 + k}\int_{0}^{\pi / 2}\dfrac{d\phi}{\sqrt{1 - \dfrac{4k}{(1 + k)^{2}}\sin^{2}\phi}}\notag\\ &= \frac{1}{1 + k}\int_{0}^{\pi / 2}\dfrac{d\phi}{\sqrt{1 - \left(\dfrac{2\sqrt{k}}{1 + k}\right)^{2}\sin^{2}\phi}}\notag\\ &= \frac{1}{1 + k}K\left(\frac{2\sqrt{k}}{1 + k}\right)\notag\end{align} With the above prologue on Landen transformations it is time to apply them on the elliptic integrals of second kind. That is the goal of the next post.

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