Carlson symmetric form

In mathematics, the Carlson symmetric forms of elliptic integrals are a small canonical set of elliptic integrals to which all others may be reduced. They are a modern alternative to the Legendre forms. The Legendre forms may be expressed in terms of the Carlson forms and vice versa.

The Carlson elliptic integrals are: $$R_F(x,y,z) = \tfrac{1}{2}\int_0^\infty \frac{dt}{\sqrt{(t+x)(t+y)(t+z)}}$$ $$R_J(x,y,z,p) = \tfrac{3}{2}\int_0^\infty \frac{dt}{(t+p)\sqrt{(t+x)(t+y)(t+z)}}$$ $$R_G(x,y,z) = \tfrac{1}{4}\int_0^\infty\frac{1}{\sqrt{(t+x)(t+y)(t+z)}} \biggl(\frac{x}{t+x} + \frac{y}{t+y} + \frac{z}{t+z} \biggr) t\,dt $$ $$R_C(x,y) = R_F(x,y,y) = \tfrac{1}{2} \int_0^\infty \frac{dt}{(t+y)\sqrt{(t+x)}}$$ $$R_D(x,y,z) = R_J(x,y,z,z) = \tfrac{3}{2} \int_0^\infty \frac{dt}{ (t+z) \,\sqrt{(t+x)(t+y)(t+z)}}$$

Since $$R_C$$ and $$R_D$$ are special cases of $$R_F$$ and $$R_J$$, all elliptic integrals can ultimately be evaluated in terms of just $$R_F$$, $$R_J$$, and $$R_G$$.

The term symmetric refers to the fact that in contrast to the Legendre forms, these functions are unchanged by the exchange of certain subsets of their arguments. The value of $$R_F(x,y,z)$$ is the same for any permutation of its arguments, and the value of $$R_J(x,y,z,p)$$ is the same for any permutation of its first three arguments.

The Carlson elliptic integrals are named after Bille C. Carlson (1924-2013).

Incomplete elliptic integrals
Incomplete elliptic integrals can be calculated easily using Carlson symmetric forms:


 * $$F(\phi,k)=\sin\phi R_F\left(\cos^2\phi,1-k^2\sin^2\phi,1\right) $$


 * $$E(\phi,k)=\sin\phi R_F\left(\cos^2\phi,1-k^2\sin^2\phi,1\right) -\tfrac{1}{3}k^2\sin^3\phi R_D\left(\cos^2\phi,1-k^2\sin^2\phi,1\right)$$


 * $$\Pi(\phi,n,k)=\sin\phi R_F\left(\cos^2\phi,1-k^2\sin^2\phi,1\right)+

\tfrac{1}{3}n\sin^3\phi R_J\left(\cos^2\phi,1-k^2\sin^2\phi,1,1-n\sin^2\phi\right)$$

(Note: the above are only valid for $$-\frac{\pi}2\le\phi\le\frac{\pi}2$$ and $$0\le k^2\sin^2\phi\le1$$)

Complete elliptic integrals
Complete elliptic integrals can be calculated by substituting &phi; = $1/2$&pi;:


 * $$K(k)=R_F\left(0,1-k^2,1\right) $$


 * $$E(k)=R_F\left(0,1-k^2,1\right)-\tfrac{1}{3}k^2 R_D\left(0,1-k^2,1\right)$$


 * $$\Pi(n,k)=R_F\left(0,1-k^2,1\right)+\tfrac{1}{3}n R_J \left(0,1-k^2,1,1-n\right) $$

Special cases
When any two, or all three of the arguments of $$R_F$$ are the same, then a substitution of $$\sqrt{t + x} = u$$ renders the integrand rational. The integral can then be expressed in terms of elementary transcendental functions.


 * $$R_{C}(x,y)

= R_{F}(x,y,y) = \frac{1}{2} \int _{0}^{\infty} \frac{dt}{\sqrt{t + x} (t + y)} = \int _{\sqrt{x}}^{\infty} \frac{du}{u^{2} - x + y} = \begin{cases} \frac{\arccos \sqrt{{x}/{y}}}{\sqrt{y - x}}, & x < y \\ \frac{1}{\sqrt{y}}, & x = y \\ \frac{\operatorname{arcosh} \sqrt{{x}/{y}}}{\sqrt{x - y}}, & x > y \\ \end{cases}$$

Similarly, when at least two of the first three arguments of $$R_J$$ are the same,


 * $$R_{J}(x,y,y,p)

= 3 \int _{\sqrt{x}}^{\infty} \frac{du}{(u^{2} - x + y) (u^{2} - x + p)} = \begin{cases} \frac{3}{p - y} (R_{C}(x,y) - R_{C}(x,p)), & y \ne p \\ \frac{3}{2 (y - x)} \left( R_{C}(x,y) - \frac{1}{y} \sqrt{x}\right), & y = p \ne x \\ \frac{1}{y^{{3}/{2}}}, &y = p = x \\ \end{cases}$$

Homogeneity
By substituting in the integral definitions $$t = \kappa u$$ for any constant $$\kappa$$, it is found that


 * $$R_F\left(\kappa x,\kappa y,\kappa z\right)=\kappa^{-1/2}R_F(x,y,z)$$


 * $$R_J\left(\kappa x,\kappa y,\kappa z,\kappa p\right)=\kappa^{-3/2}R_J(x,y,z,p)$$

Duplication theorem

 * $$R_F(x,y,z)=2R_F(x+\lambda,y+\lambda,z+\lambda)=

R_F\left(\frac{x+\lambda}{4},\frac{y+\lambda}{4},\frac{z+\lambda}{4}\right),$$

where $$\lambda=\sqrt{x}\sqrt{y}+\sqrt{y}\sqrt{z}+\sqrt{z}\sqrt{x}$$.


 * $$\begin{align}R_{J}(x,y,z,p) & = 2 R_{J}(x + \lambda,y + \lambda,z + \lambda,p + \lambda) + 6 R_{C}(d^{2},d^{2} + (p - x) (p - y) (p - z)) \\

& = \frac{1}{4} R_{J}\left( \frac{x + \lambda}{4},\frac{y + \lambda}{4},\frac{z + \lambda}{4},\frac{p + \lambda}{4}\right) + 6 R_{C}(d^{2},d^{2} + (p - x) (p - y) (p - z)) \end{align}$$

where $$d = (\sqrt{p} + \sqrt{x}) (\sqrt{p} + \sqrt{y}) (\sqrt{p} + \sqrt{z})$$ and $$\lambda =\sqrt{x}\sqrt{y}+\sqrt{y}\sqrt{z}+\sqrt{z}\sqrt{x}$$

Series Expansion
In obtaining a Taylor series expansion for $$R_{F}$$ or $$R_{J}$$ it proves convenient to expand about the mean value of the several arguments. So for $$R_{F}$$, letting the mean value of the arguments be $$A = (x + y + z)/3$$, and using homogeneity, define $$\Delta x$$, $$\Delta y$$ and $$\Delta z$$ by


 * $$\begin{align}R_{F}(x,y,z) & = R_{F}(A (1 - \Delta x),A (1 - \Delta y),A (1 - \Delta z)) \\

& = \frac{1}{\sqrt{A}} R_{F}(1 - \Delta x,1 - \Delta y,1 - \Delta z) \end{align}$$

that is $$\Delta x = 1 - x/A$$ etc. The differences $$\Delta x$$, $$\Delta y$$ and $$\Delta z$$ are defined with this sign (such that they are subtracted), in order to be in agreement with Carlson's papers. Since $$R_{F}(x,y,z)$$ is symmetric under permutation of $$x$$, $$y$$ and $$z$$, it is also symmetric in the quantities $$\Delta x$$, $$\Delta y$$ and $$\Delta z$$. It follows that both the integrand of $$R_{F}$$ and its integral can be expressed as functions of the elementary symmetric polynomials in $$\Delta x$$, $$\Delta y$$ and $$\Delta z$$ which are


 * $$E_{1} = \Delta x + \Delta y + \Delta z = 0$$


 * $$E_{2} = \Delta x \Delta y + \Delta y \Delta z + \Delta z \Delta x$$


 * $$E_{3} = \Delta x \Delta y \Delta z$$

Expressing the integrand in terms of these polynomials, performing a multidimensional Taylor expansion and integrating term-by-term...


 * $$\begin{align}R_{F}(x,y,z) & = \frac{1}{2 \sqrt{A}} \int _{0}^{\infty}\frac{1}{\sqrt{(t + 1)^{3} - (t + 1)^{2} E_{1} + (t + 1) E_{2} - E_{3}}} dt \\

& = \frac{1}{2 \sqrt{A}} \int _{0}^{\infty}\left( \frac{1}{(t + 1)^{\frac{3}{2}}} - \frac{E_{2}}{2 (t + 1)^{\frac{7}{2}}} + \frac{E_{3}}{2 (t + 1)^{\frac{9}{2}}} + \frac{3 E_{2}^{2}}{8 (t + 1)^{\frac{11}{2}}} - \frac{3 E_{2} E_{3}}{4 (t + 1)^{\frac{13}{2}}} + O(E_{1}) + O(\Delta^{6})\right) dt \\ & = \frac{1}{\sqrt{A}} \left( 1 - \frac{1}{10} E_{2} + \frac{1}{14} E_{3} + \frac{1}{24} E_{2}^{2} - \frac{3}{44} E_{2} E_{3} + O(E_{1}) + O(\Delta^{6})\right) \end{align}$$

The advantage of expanding about the mean value of the arguments is now apparent; it reduces $$E_{1}$$ identically to zero, and so eliminates all terms involving $$E_{1}$$ - which otherwise would be the most numerous.

An ascending series for $$R_{J}$$ may be found in a similar way. There is a slight difficulty because $$R_{J}$$ is not fully symmetric; its dependence on its fourth argument, $$p$$, is different from its dependence on $$x$$, $$y$$ and $$z$$. This is overcome by treating $$R_{J}$$ as a fully symmetric function of five arguments, two of which happen to have the same value $$p$$. The mean value of the arguments is therefore taken to be


 * $$A = \frac{x + y + z + 2 p}{5}$$

and the differences $$\Delta x$$, $$\Delta y$$ $$\Delta z$$ and $$\Delta p$$ defined by


 * $$\begin{align}R_{J}(x,y,z,p) & = R_{J}(A (1 - \Delta x),A (1 - \Delta y),A (1 - \Delta z),A (1 - \Delta p)) \\

& = \frac{1}{A^{\frac{3}{2}}} R_{J}(1 - \Delta x,1 - \Delta y,1 - \Delta z,1 - \Delta p) \end{align}$$

The elementary symmetric polynomials in $$\Delta x$$, $$\Delta y$$, $$\Delta z$$, $$\Delta p$$ and (again) $$\Delta p$$ are in full


 * $$E_{1} = \Delta x + \Delta y + \Delta z + 2 \Delta p = 0$$


 * $$E_{2} = \Delta x \Delta y + \Delta y \Delta z + 2 \Delta z \Delta p + \Delta p^{2} + 2 \Delta p \Delta x + \Delta x \Delta z + 2 \Delta y \Delta p$$


 * $$E_{3} = \Delta z \Delta p^{2} + \Delta x \Delta p^{2} + 2 \Delta x \Delta y \Delta p + \Delta x \Delta y \Delta z + 2 \Delta y \Delta z \Delta p + \Delta y \Delta p^{2} + 2 \Delta x \Delta z \Delta p$$


 * $$E_{4} = \Delta y \Delta z \Delta p^{2} + \Delta x \Delta z \Delta p^{2} + \Delta x \Delta y \Delta p^{2} + 2 \Delta x \Delta y \Delta z \Delta p$$


 * $$E_{5} = \Delta x \Delta y \Delta z \Delta p^{2}$$

However, it is possible to simplify the formulae for $$E_{2}$$, $$E_{3}$$ and $$E_{4}$$ using the fact that $$E_{1} = 0$$. Expressing the integrand in terms of these polynomials, performing a multidimensional Taylor expansion and integrating term-by-term as before...


 * $$\begin{align}R_{J}(x,y,z,p) & = \frac{3}{2 A^{\frac{3}{2}}} \int _{0}^{\infty}\frac{1}{\sqrt{(t + 1)^{5} - (t + 1)^{4} E_{1} + (t + 1)^{3} E_{2} - (t + 1)^{2} E_{3} + (t + 1) E_{4} - E_{5}}} dt \\

& = \frac{3}{2 A^{\frac{3}{2}}} \int _{0}^{\infty}\left( \frac{1}{(t + 1)^{\frac{5}{2}}} - \frac{E_{2}}{2 (t + 1)^{\frac{9}{2}}} + \frac{E_{3}}{2 (t + 1)^{\frac{11}{2}}} + \frac{3 E_{2}^{2} - 4 E_{4}}{8 (t + 1)^{\frac{13}{2}}} + \frac{2 E_{5} - 3 E_{2} E_{3}}{4 (t + 1)^{\frac{15}{2}}} + O(E_{1}) + O(\Delta^{6})\right) dt \\ & = \frac{1}{A^{\frac{3}{2}}} \left( 1 - \frac{3}{14} E_{2} + \frac{1}{6} E_{3} + \frac{9}{88} E_{2}^{2} - \frac{3}{22} E_{4} - \frac{9}{52} E_{2} E_{3} + \frac{3}{26} E_{5} + O(E_{1}) + O(\Delta^{6})\right) \end{align}$$

As with $$R_{J}$$, by expanding about the mean value of the arguments, more than half the terms (those involving $$E_{1}$$) are eliminated.

Negative arguments
In general, the arguments x, y, z of Carlson's integrals may not be real and negative, as this would place a branch point on the path of integration, making the integral ambiguous. However, if the second argument of $$R_C$$, or the fourth argument, p, of $$R_J$$ is negative, then this results in a simple pole on the path of integration. In these cases the Cauchy principal value (finite part) of the integrals may be of interest; these are


 * $$\mathrm{p.v.}\; R_C(x, -y) = \sqrt{\frac{x}{x + y}}\,R_C(x + y, y),$$

and


 * $$\begin{align}\mathrm{p.v.}\; R_{J}(x,y,z,-p) & = \frac{(q - y) R_{J}(x,y,z,q) - 3 R_{F}(x,y,z) + 3 \sqrt{y} R_{C}(x z,- p q)}{y + p} \\

& = \frac{(q - y) R_{J}(x,y,z,q) - 3 R_{F}(x,y,z) + 3 \sqrt{\frac{x y z}{x z + p q}} R_{C}(x z + p q,p q)}{y + p} \end{align}$$ where


 * $$q = y + \frac{(z - y) (y - x)}{y + p}.$$

which must be greater than zero for $$R_{J}(x,y,z,q)$$ to be evaluated. This may be arranged by permuting x, y and z so that the value of y is between that of x and z.

Numerical evaluation
The duplication theorem can be used for a fast and robust evaluation of the Carlson symmetric form of elliptic integrals and therefore also for the evaluation of Legendre-form of elliptic integrals. Let us calculate $$R_F(x,y,z)$$: first, define $$x_0=x$$, $$y_0=y$$ and $$z_0=z$$. Then iterate the series


 * $$\lambda_n=\sqrt{x_n}\sqrt{y_n}+\sqrt{y_n}\sqrt{z_n}+\sqrt{z_n}\sqrt{x_n},$$
 * $$x_{n+1}=\frac{x_n+\lambda_n}{4}, y_{n+1}=\frac{y_n+\lambda_n}{4}, z_{n+1}=\frac{z_n+\lambda_n}{4}$$

until the desired precision is reached: if $$x$$, $$y$$ and $$z$$ are non-negative, all of the series will converge quickly to a given value, say, $$\mu$$. Therefore,


 * $$R_F\left(x,y,z\right)=R_F\left(\mu,\mu,\mu\right)=\mu^{-1/2}.$$

Evaluating $$R_C(x,y)$$ is much the same due to the relation


 * $$R_C\left(x,y\right)=R_F\left(x,y,y\right).$$

References and External links

 * B. C. Carlson, John L. Gustafson 'Asymptotic approximations for symmetric elliptic integrals' 1993 arXiv
 * B. C. Carlson 'Numerical Computation of Real Or Complex Elliptic Integrals' 1994 arXiv
 * B. C. Carlson 'Elliptic Integrals:Symmetric Integrals' in Chap. 19 of Digital Library of Mathematical Functions. Release date 2010-05-07. National Institute of Standards and Technology.
 * 'Profile: Bille C. Carlson' in Digital Library of Mathematical Functions. National Institute of Standards and Technology.
 * Fortran code from SLATEC for evaluating RF, RJ, RC,  RD,
 * Fortran code from SLATEC for evaluating RF, RJ, RC,  RD,