Carlson symmetric form

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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: [math]\displaystyle{ R_F(x,y,z) = \tfrac{1}{2}\int_0^\infty \frac{dt}{\sqrt{(t+x)(t+y)(t+z)}} }[/math] [math]\displaystyle{ R_J(x,y,z,p) = \tfrac{3}{2}\int_0^\infty \frac{dt}{(t+p)\sqrt{(t+x)(t+y)(t+z)}} }[/math] [math]\displaystyle{ R_C(x,y) = R_F(x,y,y) = \tfrac{1}{2} \int_0^\infty \frac{dt}{(t+y)\sqrt{(t+x)}} }[/math] [math]\displaystyle{ 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)}} }[/math]

Since [math]\displaystyle{ R_C }[/math] and [math]\displaystyle{ R_D }[/math] are special cases of [math]\displaystyle{ R_F }[/math] and [math]\displaystyle{ R_J }[/math], all elliptic integrals can ultimately be evaluated in terms of just [math]\displaystyle{ R_F }[/math] and [math]\displaystyle{ R_J }[/math].

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 [math]\displaystyle{ R_F(x,y,z) }[/math] is the same for any permutation of its arguments, and the value of [math]\displaystyle{ R_J(x,y,z,p) }[/math] is the same for any permutation of its first three arguments.

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

Relation to the Legendre forms

Incomplete elliptic integrals

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

[math]\displaystyle{ F(\phi,k)=\sin\phi R_F\left(\cos^2\phi,1-k^2\sin^2\phi,1\right) }[/math]
[math]\displaystyle{ 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) }[/math]
[math]\displaystyle{ \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) }[/math]

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

Complete elliptic integrals

Complete elliptic integrals can be calculated by substituting φ = ​1⁄2π:

[math]\displaystyle{ K(k)=R_F\left(0,1-k^2,1\right) }[/math]
[math]\displaystyle{ 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) }[/math]
[math]\displaystyle{ \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) }[/math]

Special cases

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

[math]\displaystyle{ 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 \lt y \\ \frac{1}{\sqrt{y}}, & x = y \\ \frac{\operatorname{arcosh} \sqrt{{x}/{y}}}{\sqrt{x - y}}, & x \gt y \\ \end{cases} }[/math]

Similarly, when at least two of the first three arguments of [math]\displaystyle{ R_J }[/math] are the same,

[math]\displaystyle{ 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} }[/math]

Properties

Homogeneity

By substituting in the integral definitions [math]\displaystyle{ t = \kappa u }[/math] for any constant [math]\displaystyle{ \kappa }[/math], it is found that

[math]\displaystyle{ R_F\left(\kappa x,\kappa y,\kappa z\right)=\kappa^{-1/2}R_F(x,y,z) }[/math]
[math]\displaystyle{ R_J\left(\kappa x,\kappa y,\kappa z,\kappa p\right)=\kappa^{-3/2}R_J(x,y,z,p) }[/math]

Duplication theorem

[math]\displaystyle{ 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), }[/math]

where [math]\displaystyle{ \lambda=\sqrt{x}\sqrt{y}+\sqrt{y}\sqrt{z}+\sqrt{z}\sqrt{x} }[/math].

[math]\displaystyle{ \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} }[/math][1]

where [math]\displaystyle{ d = (\sqrt{p} + \sqrt{x}) (\sqrt{p} + \sqrt{y}) (\sqrt{p} + \sqrt{z}) }[/math] and [math]\displaystyle{ \lambda =\sqrt{x}\sqrt{y}+\sqrt{y}\sqrt{z}+\sqrt{z}\sqrt{x} }[/math]

Series Expansion

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

[math]\displaystyle{ \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} }[/math]

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

[math]\displaystyle{ E_{1} = \Delta x + \Delta y + \Delta z = 0 }[/math]
[math]\displaystyle{ E_{2} = \Delta x \Delta y + \Delta y \Delta z + \Delta z \Delta x }[/math]
[math]\displaystyle{ E_{3} = \Delta x \Delta y \Delta z }[/math]

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

[math]\displaystyle{ \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} }[/math]

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

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

[math]\displaystyle{ A = \frac{x + y + z + 2 p}{5} }[/math]

and the differences [math]\displaystyle{ \Delta x }[/math], [math]\displaystyle{ \Delta y }[/math] [math]\displaystyle{ \Delta z }[/math] and [math]\displaystyle{ \Delta p }[/math] defined by

[math]\displaystyle{ \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} }[/math]

The elementary symmetric polynomials in [math]\displaystyle{ \Delta x }[/math], [math]\displaystyle{ \Delta y }[/math], [math]\displaystyle{ \Delta z }[/math], [math]\displaystyle{ \Delta p }[/math] and (again) [math]\displaystyle{ \Delta p }[/math] are in full

[math]\displaystyle{ E_{1} = \Delta x + \Delta y + \Delta z + 2 \Delta p = 0 }[/math]
[math]\displaystyle{ 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 }[/math]
[math]\displaystyle{ 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 }[/math]
[math]\displaystyle{ 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 }[/math]
[math]\displaystyle{ E_{5} = \Delta x \Delta y \Delta z \Delta p^{2} }[/math]

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

[math]\displaystyle{ \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} }[/math]

As with [math]\displaystyle{ R_{J} }[/math], by expanding about the mean value of the arguments, more than half the terms (those involving [math]\displaystyle{ E_{1} }[/math]) 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 [math]\displaystyle{ R_C }[/math], or the fourth argument, p, of [math]\displaystyle{ R_J }[/math] 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

[math]\displaystyle{ \mathrm{p.v.}\; R_C(x, -y) = \sqrt{\frac{x}{x + y}}\,R_C(x + y, y), }[/math]

and

[math]\displaystyle{ \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} }[/math]

where

[math]\displaystyle{ q = y + \frac{(z - y) (y - x)}{y + p}. }[/math]

which must be greater than zero for [math]\displaystyle{ R_{J}(x,y,z,q) }[/math] 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 [math]\displaystyle{ R_F(x,y,z) }[/math]: first, define [math]\displaystyle{ x_0=x }[/math], [math]\displaystyle{ y_0=y }[/math] and [math]\displaystyle{ z_0=z }[/math]. Then iterate the series

[math]\displaystyle{ \lambda_n=\sqrt{x_n}\sqrt{y_n}+\sqrt{y_n}\sqrt{z_n}+\sqrt{z_n}\sqrt{x_n}, }[/math]
[math]\displaystyle{ 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} }[/math]

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

[math]\displaystyle{ R_F\left(x,y,z\right)=R_F\left(\mu,\mu,\mu\right)=\mu^{-1/2}. }[/math]

Evaluating [math]\displaystyle{ R_C(x,y) }[/math] is much the same due to the relation

[math]\displaystyle{ R_C\left(x,y\right)=R_F\left(x,y,y\right). }[/math]

References and External links

  1. ↑ Carlson, Bille C. (1994). "Numerical computation of real or complex elliptic integrals". arXiv:math/9409227v1.



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