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In [[mathematics]], the '''Carlson symmetric forms of [[elliptic integral]]s''' are a small canonical set of elliptic integrals to which all others may be reduced. They are a modern alternative to the [[Legendre form]]s. The Legendre forms may be expressed in terms of the Carlson forms and vice versa.
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The Carlson elliptic integrals are:
 
:<math>R_F(x,y,z) = \tfrac{1}{2}\int_0^\infty \frac{dt}{\sqrt{(t+x)(t+y)(t+z)}}</math>
 
:<math>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>R_C(x,y) = R_F(x,y,y) = \tfrac{1}{2} \int_0^\infty \frac{dt}{(t+y)\sqrt{(t+x)}}</math>
 
:<math>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>\scriptstyle{R_C}</math> and <math>\scriptstyle{R_D}</math> are special cases of <math>\scriptstyle{R_F}</math> and <math>\scriptstyle{R_J}</math>, all elliptic integrals can ultimately be evaluated in terms of just <math>\scriptstyle{R_F}</math> and <math>\scriptstyle{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 of their arguments. The value of <math>\scriptstyle{R_F(x,y,z)}</math> is the same for any permutation of its arguments, and the value of <math>\scriptstyle{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.
 
==Relation to the Legendre forms==
 
===Incomplete elliptic integrals===
 
Incomplete [[elliptic integral]]s can be calculated easily using Carlson symmetric forms:
 
:<math>F(\phi,k)=\sin\phi R_F\left(\cos^2\phi,1-k^2\sin^2\phi,1\right) </math>
 
:<math>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>\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>0\le\phi\le2\pi</math> and <math>0\le k^2\sin^2\phi\le1</math>)
 
===Complete elliptic integrals===
 
Complete [[elliptic integral]]s can be calculated by substituting &phi;&nbsp;=&nbsp;{{frac|1|2}}&pi;:
 
:<math>K(k)=R_F\left(0,1-k^2,1\right) </math>
 
:<math>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>\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>R_F</math> are the same, then a substitution of <math>\sqrt{t + x} = u</math> renders the integrand rational. The integral can then be expressed in terms of elementary transcendental functions.
 
:<math>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}</math>
 
Similarly, when at least two of the first three arguments of <math>R_J</math> are the same,
 
:<math>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}</math>
 
==Properties==
 
===Homogeneity===
 
By substituting in the integral definitions <math>t = \kappa u</math> for any constant <math>\kappa</math>, it is found that
 
:<math>R_F\left(\kappa x,\kappa y,\kappa z\right)=\kappa^{-1/2}R_F(x,y,z)</math>
 
:<math>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>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>\lambda=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}</math>.
 
:<math>\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>
 
where <math>d = (\sqrt{p} + \sqrt{x}) (\sqrt{p} + \sqrt{y}) (\sqrt{p} + \sqrt{z})</math> and <math>\lambda = \sqrt{x y} + \sqrt{y z} + \sqrt{z x}</math>
 
==Series Expansion==
 
In obtaining a [[Taylor series]] expansion for <math>\scriptstyle{R_{F}}</math> or <math>\scriptstyle{R_{J}}</math> it proves convenient to expand about the mean value of the several arguments. So for <math>\scriptstyle{R_{F}}</math>, letting the mean value of the arguments be <math>\scriptstyle{A = (x + y + z)/3}</math>, and using homogeneity, define <math>\scriptstyle{\Delta x}</math>, <math>\scriptstyle{\Delta y}</math> and <math>\scriptstyle{\Delta z}</math> by
 
:<math>\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>\scriptstyle{\Delta x = 1 - x/A}</math> etc. The differences <math>\scriptstyle{\Delta x}</math>, <math>\scriptstyle{\Delta y}</math> and <math>\scriptstyle{\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>\scriptstyle{R_{F}(x,y,z)}</math> is symmetric under permutation of <math>\scriptstyle{x}</math>, <math>\scriptstyle{y}</math> and <math>\scriptstyle{z}</math>, it is also symmetric in the quantities <math>\scriptstyle{\Delta x}</math>, <math>\scriptstyle{\Delta y}</math> and <math>\scriptstyle{\Delta z}</math>. It follows that both the integrand of <math>\scriptstyle{R_{F}}</math> and its integral can be expressed as functions of the [[elementary symmetric polynomial]]s in <math>\scriptstyle{\Delta x}</math>, <math>\scriptstyle{\Delta y}</math> and <math>\scriptstyle{\Delta z}</math> which are
 
:<math>E_{1} = \Delta x + \Delta y + \Delta z = 0</math>
 
:<math>E_{2} = \Delta x \Delta y + \Delta y \Delta z + \Delta z \Delta x</math>
 
:<math>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>\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>\scriptstyle{E_{1}}</math> identically to zero, and so eliminates all terms involving <math>\scriptstyle{E_{1}}</math> - which otherwise would be the most numerous.
 
An ascending series for <math>\scriptstyle{R_{J}}</math> may be found in a similar way. There is a slight difficulty because <math>\scriptstyle{R_{J}}</math> is not fully symmetric; its dependence on its fourth argument, <math>\scriptstyle{p}</math>, is different from its dependence on <math>\scriptstyle{x}</math>, <math>\scriptstyle{y}</math> and <math>\scriptstyle{z}</math>. This is overcome by treating <math>\scriptstyle{R_{J}}</math> as a fully symmetric function of ''five'' arguments, two of which happen to have the same value <math>\scriptstyle{p}</math>. The mean value of the arguments is therefore take to be
 
:<math>A = \frac{x + y + z + 2 p}{5}</math>
 
and the differences <math>\scriptstyle{\Delta x}</math>, <math>\scriptstyle{\Delta y}</math> <math>\scriptstyle{\Delta z}</math> and <math>\scriptstyle{\Delta p}</math> defined by
 
:<math>\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 polynomial]]s in <math>\scriptstyle{\Delta x}</math>, <math>\scriptstyle{\Delta y}</math>, <math>\scriptstyle{\Delta z}</math>, <math>\scriptstyle{\Delta p}</math> and (again) <math>\scriptstyle{\Delta p}</math> are in full
 
:<math>E_{1} = \Delta x + \Delta y + \Delta z + 2 \Delta p = 0</math>
 
:<math>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>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>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>E_{5} = \Delta x \Delta y \Delta z \Delta p^{2}</math>
 
However, it is possible to simplify the formulae for <math>\scriptstyle{E_{2}}</math>, <math>\scriptstyle{E_{3}}</math> and <math>\scriptstyle{E_{4}}</math> using the fact that <math>\scriptstyle{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>\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>\scriptstyle{R_{J}}</math>, by expanding about the mean value of the arguments, more than half the terms (those involving <math>\scriptstyle{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>\scriptstyle{R_C}</math>, or the fourth argument, p, of <math>\scriptstyle{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>\mathrm{p.v.}\; R_C(x, -y) = \sqrt{\frac{x}{x + y}}\,R_C(x + y, y),</math>
 
and
 
:<math>\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>q = y + \frac{(z - y) (y - x)}{y + p}.</math>
 
which must be greater than zero for <math>\scriptstyle{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>R_F(x,y,z)</math>:
first, define <math>x_0=x</math>, <math>y_0=y</math> and <math>z_0=z</math>. Then iterate the series
 
:<math>\lambda_n=\sqrt{x_ny_n}+\sqrt{y_nz_n}+\sqrt{z_nx_n},</math>
:<math>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>x</math>, <math>y</math> and <math>z</math> are non-negative, all of the series will converge quickly to a given value, say, <math>\mu</math>. Therefore,
 
:<math>R_F\left(x,y,z\right)=R_F\left(\mu,\mu,\mu\right)=\mu^{-1/2}.</math>
 
Evaluating <math>R_C(x,y)</math> is much the same due to the relation
 
:<math>R_C\left(x,y\right)=R_F\left(x,y,y\right).</math>
 
==References and External links==
*[http://arxiv.org/abs/math/9310223v1 B. C. Carlson, John L. Gustafson 'Asymptotic approximations for symmetric elliptic integrals' 1993 arXiv]
*[http://arxiv.org/abs/math/9409227v1 B. C. Carlson 'Numerical Computation of Real Or Complex Elliptic Integrals' 1994 arXiv]
*[http://dlmf.nist.gov/19#PT3 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.]
*[http://dlmf.nist.gov/about/bio/BCCarlson 'Profile: Bille C. Carlson' in ''Digital Library of Mathematical Functions''. National Institute of Standards and Technology.]
 
*{{Citation | last1=Press | first1=WH | last2=Teukolsky | first2=SA | last3=Vetterling | first3=WT | last4=Flannery | first4=BP | year=2007 | title=Numerical Recipes: The Art of Scientific Computing | edition=3rd | publisher=Cambridge University Press |  publication-place=New York | isbn=978-0-521-88068-8 | chapter=Section 6.12. Elliptic Integrals and Jacobian Elliptic Functions | chapter-url=http://apps.nrbook.com/empanel/index.html#pg=309}}
 
*[[Fortran]] code from [[SLATEC]] for evaluating [http://www.netlib.org/slatec/src/drf.f RF],  [http://www.netlib.org/slatec/src/drj.f RJ], [http://www.netlib.org/slatec/src/drc.f RC],  [http://www.netlib.org/slatec/src/drd.f RD],
 
[[Category:Elliptic functions]]

Latest revision as of 18:30, 10 August 2014

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