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:

$R_F(x,y,z) = \frac{1}{2}\int_0^\infty \frac{dt}{\sqrt{(t+x)(t+y)(t+z)}}$

$R_J(x,y,z,p) = \frac{3}{2}\int_0^\infty \frac{dt}{(t+p)\sqrt{(t+x)(t+y)(t+z)}}$

$R_C(x,y) = R_F(x,y,y) = \frac{1}{2} \int_0^\infty \frac{dt}{(t+y)\sqrt{(t+x)}}$

$R_D(x,y,z) = R_J(x,y,z,z) = \frac{3}{2} \int_0^\infty \frac{dt}{ (t+z) \,\sqrt{(t+x)(t+y)(t+z)}}$

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

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

Relation to the Legendre forms

Incomplete elliptic integrals

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

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

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

$\Pi(\phi,n,k)=\sin\phi R_F\left(\cos^2\phi,1-k^2\sin^2\phi,1\right)+ \frac{1}{3}n\sin^3\phi R_J\left(\cos^2\phi,1-k^2\sin^2\phi,1,1-n\sin^2\phi\right)$

(Note: the above are only valid for $0\le\phi\le2\pi$ and $0\le k^2\sin^2\phi\le1$)

Complete elliptic integrals

Complete elliptic integrals can be calculated by substituting $\phi=\frac{\pi}{2}$:

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

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

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

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.

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

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

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

Properties

Homogeneity

By substituting in the integral definitions t = κu for any constant κ, 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 $\lambda=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}$.

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

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

Series Expansion

In obtaining a Taylor series expansion for $\scriptstyle{R_{F}}$ or $\scriptstyle{R_{J}}$ it proves convenient to expand about the mean value of the several arguments. So for $\scriptstyle{R_{F}}$, letting the mean value of the arguments be $\scriptstyle{A = (x + y + z)/3}$, and using homogeneity, define $\scriptstyle{\Delta x}$, $\scriptstyle{\Delta y}$ and $\scriptstyle{\Delta z}$ 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 $\scriptstyle{\Delta x = 1 - x/A}$ etc. The differences $\scriptstyle{\Delta x}$, $\scriptstyle{\Delta y}$ and $\scriptstyle{\Delta z}$ are defined with this sign (such that they are subtracted), in order to be in agreement with Carlson's papers. Since $\scriptstyle{R_{F}(x,y,z)}$ is symmetric under permutation of $\scriptstyle{x}$, $\scriptstyle{y}$ and $\scriptstyle{z}$, it is also symmetric in the quantities $\scriptstyle{\Delta x}$, $\scriptstyle{\Delta y}$ and $\scriptstyle{\Delta z}$. It follows that both the integrand of $\scriptstyle{R_{F}}$ and its integral can be expressed as functions of the elementary symmetric polynomials in $\scriptstyle{\Delta x}$, $\scriptstyle{\Delta y}$ and $\scriptstyle{\Delta z}$ which are

E₁ = Δx + Δy + Δz = 0

E₂ = ΔxΔy + ΔyΔz + ΔzΔx

E₃ = ΔxΔyΔz

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 $\scriptstyle{E_{1}}$ identically to zero, and so eliminates all terms involving $\scriptstyle{E_{1}}$ - which otherwise would be the most numerous.

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

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

and the differences $\scriptstyle{\Delta x}$, $\scriptstyle{\Delta y}$ $\scriptstyle{\Delta z}$ and $\scriptstyle{\Delta p}$ 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 $\scriptstyle{\Delta x}$, $\scriptstyle{\Delta y}$, $\scriptstyle{\Delta z}$, $\scriptstyle{\Delta p}$ and (again) $\scriptstyle{\Delta p}$ are in full

E₁ = Δx + Δy + Δz + 2Δp = 0

E₂ = ΔxΔy + ΔyΔz + 2ΔzΔp + Δp² + 2ΔpΔx + ΔxΔz + 2ΔyΔp

E₃ = ΔzΔp² + ΔxΔp² + 2ΔxΔyΔp + ΔxΔyΔz + 2ΔyΔzΔp + ΔyΔp² + 2ΔxΔzΔp

E₄ = ΔyΔzΔp² + ΔxΔzΔp² + ΔxΔyΔp² + 2ΔxΔyΔzΔp

E₅ = ΔxΔyΔzΔp²

However, it is possible to simplify the formulae for $\scriptstyle{E_{2}}$, $\scriptstyle{E_{3}}$ and $\scriptstyle{E_{4}}$ using the fact that $\scriptstyle{E_{1} = 0}$. 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 $\scriptstyle{R_{J}}$, by expanding about the mean value of the arguments, more than half the terms (those involving $\scriptstyle{E_{1}}$) 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 $\scriptstyle{R_C}$, or the fourth argument, p, of $\scriptstyle{R_J}$ 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 $\scriptstyle{R_{J}(x,y,z,q)}$ 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 x₀ = x, y₀ = y and z₀ = z. Then iterate the series

$\lambda_n=\sqrt{x_ny_n}+\sqrt{y_nz_n}+\sqrt{z_nx_n},$

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

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, μ. 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.

• Press, WH; Teukolsky, SA; Vetterling, WT; Flannery, BP (2007), "Section 6.12. Elliptic Integrals and Jacobian Elliptic Functions", Numerical Recipes: The Art of Scientific Computing (3rd ed.), New York: Cambridge University Press, ISBN 978-0-521-88068-8, http://apps.nrbook.com/empanel/index.html#pg=309