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


*

*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.

*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 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 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 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
A: Short answer: yes you can. After accepting and upvoting Sandeep's question, however, the algorithm I settled on was not the algorithm he proposed.
I managed (through a friend's assistance) to see the first two chapters of Borwein and Borwein's "Pi and the AGM". The identity that I needed, it turns out, was exactly the algorithm for the nome $q$ that I had mentioned in my comment to his answer (i.e., a hijacked AGM iteration that computes both $q=\exp(-\pi K(1-m)/K(m))$ and $K(m)$ simultaneously). It is equation 2.5.14 in "Pi and the AGM":
$q=\frac{m}{16}\prod_{j=0}^\infty{\left(\frac{a_j}{a_{j+1}}\right)^{2^{1-j}}}$
where $a_j$ is an iterate from the AGM iteration.
In pseudocode, here is the algorithm I now have for computing $K(m)$, $E(m)$, and $t=\ln(q(m))$ simultaneously:
a←(1+√(1-m))/2
c←m/(4a)
t←ln(c/(4a))
s←a^2
f←1
repeat
v←(a+√((a-c)(a+c)))/2
t←t+ln(a/v)/f
a←v
c←(c^2)/(4a)
f←2f
s←s-fc^2
until |c| is small enough
K←π/(2a)
E←KS

The code fails for $m=0$ and $m=1$, but these trivial cases can be handled separately. $K(1-m)$ is then -Kt/π. $E(1-m)$ is of course then computed through the Legendre relation.
