# Inverse of elliptic integral of second kind

The Wikipedia articles on elliptic integral and elliptic functions state that “elliptic functions were discovered as inverse functions of elliptic integrals.” Some elliptic functions have names and are thus well-known special functions, and the same holds for some elliptic integrals. But what is the relation between the named elliptic functions and the named elliptic integrals?

It seems that the Jacobi amplitude $\varphi=\operatorname{am}(u,k)$ is the inverse of the elliptic integral of the first kind, $u=F(\varphi,k)$. Or related to this, $x=\operatorname{sn}(u,k)$ is the inverse of $u=F(x;k)$. It looks to me as if all of Jacobi's elliptic functions relate to the elliptic integral of the first kind. For other named elliptic functions listed by Wikipedia, like Jacobi's $\vartheta$ function or Weierstrass's $\wp$ function, it is even harder to see a relation to Legendre's integrals.

Is there a way to express the inverse of $E$, the elliptic integral of the second kind, in terms of some named elliptic functions? I.e. given $E(\varphi,k)=u$, can you write a closed form expression for $\varphi$ in terms of $k$ and $u$ using well-known special functions and elementary arithmetic operations?

In this post the author uses the Mathematica function FindRoot to do this kind of inversion, but while reading that post, I couldn't help wondering whether there is an easier formulation. Even though the computation behind the scenes might in fact boil down to root-finding in any case, it feels like this task should be common enough that someone has come up with a name for the core of this computation.

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• This paper might be interesting for you: Numerical computation of incomplete elliptic integrals of a general form by T. Fukushima and H. Ishizaki. Also the paper Numerical computation of inverse complete elliptic integrals of first and second kinds by T Fukushima. – Loreno Heer Jan 28 '15 at 16:07
• yes inverse calculations are in the second paper – Loreno Heer Jan 28 '15 at 16:08
• @sanjab: Thanks for these pointers. But as far as I can see, the first paper only does forward computation, and the second does inverse (using notation like $m_E$), but of the complete integral and solving for the parameter $m$ not the amplitude $\varphi$. So at least as far as I can see, this is related but won't immediately help answering my question. – MvG Jan 28 '15 at 16:29
• – J. M. is a poor mathematician May 17 '16 at 3:40
• N.B. the inverse of the Weierstrass elliptic function can of course be expressed in terms of $F$, or Carlson's $R_F$ if you prefer the symmetric integrals. Similarly, the Weierstrass and Jacobi functions are expressible in terms of each other (corresponding to the algebro-geometric notion of a quartic being equivalent to a cubic under suitable rational transformations.) – J. M. is a poor mathematician May 17 '16 at 3:43

I've just found a physical problem (in classical mechanics) involving the trajectory of a particle in which I had to take the inverse of $E(\phi,k)$, the incomplete elliptic integral of the 2nd kind. For the elliptic integral of the 1st kind $F(x,k)$ this is an easy task beacuse the Jacobi elliptic function $sn(x,k)$ (or JacobiSN(x,k) in mathematical software) is just $F^{-1}(x,k)$. However, currently there is no built-in function for the inverse of $E(\phi,k)$. My computational solution was then to build a procedure for the inverse, using FindRoot (in Mathematica9.0) or fsolve (in Maple 2015). F.M.S. Lima (University of Brasilia).
Write $$E(\phi,k) = \sum_{i=0}^\infty \frac{Q_i(k)}{(2i-1)!}\phi^{2i-1}$$ where $Q_i(k)$ are polynomials in $k$, from the series expansion here we can get a finite form for these polynomials as $$Q_n(m) = 2(-4)^n\sum_{k=1}^n \frac{(2k-3)!!}{k!}\left(\frac{-m}{8}\right)^k \sum_{j=0}^{k-1} \binom{2k}{j}(-1)^{1-j}(j-k)^{2n}, \;\;\; n>0$$ with $Q_0(k)$ defined as $1$. Then you can write the inverse series using series reversion in a very similar manner to $E(\phi,k)$ $$\phi(E,k) = \sum_{i=0}^\infty \frac{R_i(k)}{(2i-1)!}E^{2i-1}$$ where the relation between the new polynomials $R_i(k)$ is given by the explicit reversion formula found at the bottom of this link, giving $$R_n(k) = (2n)! \sum_{\tau_n=n}(-1)^{\sigma_n} \frac{\prod_{i=1}^{\sigma_n}2n+i}{\prod_{j=1}^n k_j!}\prod_{l=1}^n \left(\frac{Q_l(k)}{(2l+1)!}\right)^{k_l}$$ where $\sigma_n=k_1+k_2+k_3+\cdots+k_n$, and $\tau_n=k_1+2k_2+3k+3+\cdots + nk_n$ and the sum is over all sets of indices $k_i$ that meet the requirement $\tau_n=n$. I don't know if any nice simplifications or tricks can be made to reduce this to a functional form. The only numerical element here is converging the series to the desired accuracy.