On the elliptic logarithm and the Carlson symmetric elliptic integral

August 16, 2010

Nothing too deep for this blog entry, but since neither DLMF nor Wolfram Functions has this listed, I’ll go ahead and save it here for posterity.

Wolfram MathWorld defines the “elliptic logarithm” as follows (rearranged a bit for clarity):

\mathrm{eln}_{a,b}\left(z\right)=-\frac1{2}\int_{z}^{\infty}\frac{\mathrm{d}y}{\sqrt{y^3+a y^2+b y}}

which resembles the defining expression for the Carlson symmetric elliptic integral of the first kind R_F:


To segue a bit, as noted in the MathWorld entry, it is rather annoying that the corresponding function EllipticLog[] in Mathematica has a superfluous second parameter that merely serves to change the sign of the output in the real case.

However, I note here that the phrase “complicated form involving incomplete elliptic integrals of the first kind with complex parameters” in the MathWorld entry is true only if one sticks to using the Legendre incomplete elliptic integral of the first kind F(\phi|m), as shown in this Wolfram Functions entry. If one uses the Carlson form instead, the expression is only a bit more complicated.

After making the substitution y=t+z in the defining integral for the elliptic logarithm, one simply factors the denominator expression into


where v_1 and v_2 are the roots of the polynomial v^2+av+b.

At this point, you the reader might ask why I chose not to use the quadratic formula directly here. The problem is that naïvely applying the formula


can result in one of the roots being computed inaccurately whenever b is tiny relative to a (computing one of the roots then requires the subtraction of two nearly equal numbers, which is always a bad thing to do in numerical work). The right way to go about it is to choose the sign of the square root such that \Re\left(\bar{a}\sqrt{a^2-4b}\right)\geq 0, after which the two roots are

v_1=-\frac{a+\sqrt{a^2-4b}}{2} and v_2=-\frac{2b}{a+\sqrt{a^2-4b}}

Thus, the elliptic logarithm and the Carlson symmetric elliptic integral have the following relation:


where of course the three arguments of R_F can be permuted as appropriate due to symmetry.

Since there are published algorithms for computing the Carlson integrals, this is how one might compute the elliptic logarithm when one has the Carlson functions available.

The only limitation for this relation is that R_F is only defined when all of its arguments lie in the complex plane cut along the negative real axis; thus, for instance, one cannot use this relation when z is a negative real. One might be able to use the homogeneity relation for R_F, but I still have to investigate the proper way of going about it.



questions, questions

August 13, 2010

I really should be getting around to posting something with actual mathematical content in this blog, but for now I’m slowly working through my backlog of my “dumb questions”; and there were quite a lot of them scattered around in my handwritten notes and an assorted bunch of my Mathematica notebook experiments. Thus far, my technique of asking my dumb questions at MathOverflow (and to a lesser extent at “Math Underflow“) seems to be working: if they get downvoted, then they’re definitely dumb.

In at least two separate occasions, some of the answers I got led me to useful new ideas, though not necessarily related to the original question(s) I asked. I guess that’s something. :)

Once I sort myself out, I’ll start posting about my experiments.

Happy trails!