The complete elliptic integral of the first kind


$K(m)=\int\_0^{\pi/2}\frac{\mathrm{d}t}{\sqrt{1-m\sin^2t}}$


is easily computed via the arithmetic-geometric mean iteration; to wit,


$K(m)=\frac{\pi}{2M(1,\sqrt{1-m})}$


where $M(a,b)$ is the arithmetic-geometric mean of $a$ and $b$. With a little more trickery, the iteration can be hijacked to compute the complete elliptic integral of the second kind $E(m)$ as well.


In a number of applications, it happens that one needs both the values of $K(m)$ and its complement $K(1-m)$ (and sometimes similarly for $E(m)$ and $E(1-m)$).


My question is, apart from having to do an AGM iteration for each of $K(m)$ and $K(1-m)$, is there an algorithm (maybe a modification of the basic AGM iteration) that simultaneously generates both $K(m)$ and its complement? I would also be interested in seeing also an extension of this algorithm, if one exists, for computing $E(m)$ as well (after which $E(1-m)$ is easily computed via Legendre's relation).



Simultaneously computing a complete elliptic integral and its complement
There are two possible ways to attack this problem


1. Both K and K' can be expressed in
terms of the Theta function as
described here
<http://mathworld.wolfram.com/EllipticModulus.html>.
If you compute $\Theta\_3$, you can get
both at the same time.
2. The other way is to observe that
both K and K' are expressible in terms of the hypergeometric function $\_2F\_1(\frac{1}{2}, \frac{1}{2} ; 1; m)$. They are solutions of the
same self-adjoint Gauss
hypergeometric differential
equation (since the equation is invariant under the transformation (m $\rightarrow$ 1-m))


$(k^3 - k)\frac{d^2y}{dk^2} + (3k^2
 -1)\frac{dy}{dk} + ky = 0 $ 


By virtue of this fact, both K and K' are connected. You will find the following series expansion for K'(k) 
derived in [Borwein's book Pi and
AGM](http://rads.stackoverflow.com/amzn/click/047131515X) Section 1.3


$ K'(k) = \frac{2}{\pi} log \frac
 {4}{k} K(k) - 2 [(\frac{1}{2})^2(\frac{1}{1.2}k^2 + (\frac{1.3}{2.4})^2(\frac{1}{1.2} + \frac{1}{3.4})k^4 + (\frac{1.3.5}{2.4.6})^2(\frac{1}{1.2} + \frac{1}{3.4} + \frac{1}{5.6})k^6 $.....(infinite series) + ]


You may also find Chapter 5 of [Armitage and Eberlein's book on Elliptic Functions](http://rads.stackoverflow.com/amzn/click/0521780780) useful.


EDIT1: I put in the complete series expansion for K'(k). 




---


Regarding the computation of E(k), E(k) and K(k) are connected by the differential equation $ \frac{dK}{dk} = \frac{E - (1-k^2)K}{k(1-k^2)} $ which is how the [Legendre relation](http://mathworld.wolfram.com/LegendreRelation.html) you mention above comes about. 


Again Borwein has the solution for this problem(buy the book!). Exercise 3 in Sec 1.4 has the formula based on the quartic AGM iteration
$ E(k) = K(k)[1 - \sum\_{n=0}^{\infty} 4^n [\alpha\_n^4 - (\frac{\alpha\_n^2+\beta\_n^2}{2})^2 ] $ where


* $\alpha\_n = (a\_{2n})^{\frac{1}{2}}
 and \beta\_n = (b\_{2n})^{\frac{1}{2}}$


and $ a\_n, b\_n$ and $c\_n $ satisfy the AGM relation


