Symmetrical Carlson shape

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

The elliptical Carlson integrals are:

${\displaystyle R_{F}(x,y,z)={\tfrac {1}{2}}\int _{0}^{\infty }{\frac {\mathrm {d} t}{\sqrt {(t+x)(t+y)(t+z)}}}}$

${\displaystyle R_{J}(x,y,z,p)={\tfrac {3}{2}}\int _{0}^{\infty }{\frac {\mathrm {d} t}{(t+p){\sqrt {(t+x)(t+y)(t+z)}}}}}$

${\displaystyle R_{C}(x,y)=R_{F}(x,y,y)={\tfrac {1}{2}}\int _{0}^{\infty }{\frac {\mathrm {d} t}{(t+y){\sqrt {(t+x)}}}}}$

${\displaystyle R_{D}(x,y,z)=R_{J}(x,y,z,z)={\tfrac {3}{2}}\int _{0}^{\infty }{\frac {\mathrm {d} t}{(t+z)\,{\sqrt {(t+x)(t+y)(t+z)}}}}}$

Since and are special cases of and , all elliptic integrals can ultimately be represented by and . ${\displaystyle R_{C}}$ ${\displaystyle R_{D}}$ ${\displaystyle R_{F}}$ ${\displaystyle R_{J}}$ ${\displaystyle R_{F}}$ ${\displaystyle R_{J}}$

The term symmetric refers to the fact that, unlike the Legendre forms, these functions remain unchanged by interchanging certain function arguments. The value of is the same for each permutation of the arguments, and the value of is the same for each permutation of the first three arguments. ${\displaystyle R_{F}(x,y,z)}$ ${\displaystyle R_{J}(x,y,z,p)}$

The elliptical Carlson integrals are named after Bille C. Carlson.

Connection with the Legendre forms

Incomplete elliptic integrals

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

${\displaystyle F(\phi ,k)=\sin \phi \;R_{F}({\cos ^{2}}\phi ,1-{k^{2}\sin ^{2}}\phi ,1)}$

${\displaystyle E(\phi ,k)=\sin \phi \;R_{F}({\cos ^{2}}\phi ,1-{k^{2}\sin ^{2}}\phi ,1)-{\tfrac {1}{3}}{k^{2}\sin ^{3}}\phi \;R_{D}({\cos ^{2}}\phi ,1-{k^{2}\sin ^{2}}\phi ,1)}$

${\displaystyle \Pi (\phi ,n,k)=\sin \phi \;R_{F}({\cos ^{2}}\phi ,1-{k^{2}\sin ^{2}}\phi ,1)+{\tfrac {1}{3}}{n\sin ^{3}}\phi \;R_{J}({\cos ^{2}}\phi ,1-{k^{2}\sin ^{2}}\phi ,1,1-{n\sin ^{2}}\phi )}$

(Note: this only applies to and ) ${\displaystyle 0\leq \phi \leq 2\pi }$ ${\displaystyle 0\leq k^{2}\sin ^{2}\phi \leq 1}$

Complete elliptic integrals

Complete elliptic integrals can be calculated by substituting φ = π / 2:

${\displaystyle K(k)=R_{F}\left(0,1-k^{2},1\right)}$

${\displaystyle E(k)=R_{F}\left(0,1-k^{2},1\right)-{\tfrac {1}{3}}k^{2}R_{D}\left(0,1-k^{2},1\right)}$

${\displaystyle \Pi (n,k)=R_{F}\left(0,1-k^{2},1\right)+{\tfrac {1}{3}}nR_{J}\left(0,1-k^{2},1,1-n\right)}$

Special cases

If any two or all three arguments of are identical, then the substitution makes the integrand rational. The integral can then be expressed in terms of elementary transcendent functions . ${\displaystyle R_{F}}$ ${\displaystyle {\sqrt {t+x}}=u}$

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

The situation is similar if at least two of the first three arguments of are identical, ${\displaystyle R_{J}}$

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

properties

homogeneity

If you replace every constant with in the integral definitions , you find that ${\displaystyle \kappa }$ ${\displaystyle t=\kappa u}$

${\displaystyle R_{F}\left(\kappa x,\kappa y,\kappa z\right)=\kappa ^{-1/2}R_{F}(x,y,z)}$

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

Duplication set

${\displaystyle R_{F}(x,y,z)=2R_{F}(x+\lambda ,y+\lambda ,z+\lambda )=R_{F}\left({\frac {x+\lambda }{4}},{\frac {y+\lambda }{4}},{\frac {z+\lambda }{4}}\right),}$

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

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

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

Series development

To get a Taylor series for or , it comes in handy to develop the mean of all the arguments. For the average of the arguments is so , and using the homogeneity are , and defined by ${\displaystyle R_{F}}$ ${\displaystyle R_{J}}$ ${\displaystyle R_{F}}$ ${\displaystyle A=(x+y+z)/3}$ ${\displaystyle \Delta x}$ ${\displaystyle \Delta y}$ ${\displaystyle \Delta z}$

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

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

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

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

${\displaystyle E_{3}=\Delta x\Delta y\Delta z}$ .

Expressed the integrand by these polynomials, carried out a multi-dimensional Taylor expansion and integrated term by term, results

${\displaystyle {\begin{aligned}R_{F}(x,y,z)&={\frac {1}{2{\sqrt {A}}}}\int _{0}^{\infty }{\frac {1}{\sqrt {(t+1)^{3}-(t+1)^{2}E_{1}+(t+1)E_{2}-E_{3}}}}\,\mathrm {d} t\\&={\frac {1}{2{\sqrt {A}}}}\int _{0}^{\infty }\left({\frac {1}{(t+1)^{3/2}}}-{\frac {E_{2}}{2(t+1)^{7/2}}}+{\frac {E_{3}}{2(t+1)^{9/2}}}+{\frac {3E_{2}^{2}}{8(t+1)^{11/2}}}-{\frac {3E_{2}E_{3}}{4(t+1)^{13/2}}}+O(E_{1})+O(\Delta ^{6})\right)\mathrm {d} t\\&={\frac {1}{\sqrt {A}}}\left(1-{\frac {1}{10}}E_{2}+{\frac {1}{14}}E_{3}+{\frac {1}{24}}E_{2}^{2}-{\frac {3}{44}}E_{2}E_{3}+O(E_{1})+O(\Delta ^{6})\right)\end{aligned}}}$

The advantage of developing around the mean of the arguments is now revealed; it reduces to zero and thus eliminates all terms that would otherwise be the most numerous. ${\displaystyle E_{1}}$ ${\displaystyle E_{1}}$

An ascending row for can be found in a similar way. There is a little difficulty because it is not completely symmetrical; its dependence on the fourth argument,, is different from the dependence on , and . This is overcome by treating as a fully symmetric function of five arguments, two of which now have the same value . The mean of the arguments will therefore be ${\displaystyle R_{J}}$ ${\displaystyle R_{J}}$ ${\displaystyle p}$ ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle z}$ ${\displaystyle R_{J}}$ ${\displaystyle p}$

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

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

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

The elementary symmetric polynomials of , , , and (again) total ${\displaystyle \Delta x}$ ${\displaystyle \Delta y}$ ${\displaystyle \Delta z}$ ${\displaystyle \Delta p}$ ${\displaystyle \Delta p}$

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

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

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

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

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

However, it is possible to simplify the formulas for , and using the fact that . Expressed the integrand by these polynomials, performed a multi-dimensional Taylor expansion and integrated term by term, as before ${\displaystyle E_{2}}$ ${\displaystyle E_{3}}$ ${\displaystyle E_{4}}$ ${\displaystyle E_{1}=0}$

${\displaystyle {\begin{aligned}R_{J}(x,y,z,p)&={\frac {3}{2A^{3/2}}}\int _{0}^{\infty }{\frac {1}{\sqrt {(t+1)^{5}-(t+1)^{4}E_{1}+(t+1)^{3}E_{2}-(t+1)^{2}E_{3}+(t+1)E_{4}-E_{5}}}}\,\mathrm {d} t\\&={\frac {3}{2A^{3/2}}}\int _{0}^{\infty }\left({\frac {1}{(t+1)^{5/2}}}-{\frac {E_{2}}{2(t+1)^{9/2}}}+{\frac {E_{3}}{2(t+1)^{11/2}}}+{\frac {3E_{2}^{2}-4E_{4}}{8(t+1)^{13/2}}}+{\frac {2E_{5}-3E_{2}E_{3}}{4(t+1)^{15/2}}}+O(E_{1})+O(\Delta ^{6})\right)\mathrm {d} t\\&={\frac {1}{A^{3/2}}}\left(1-{\frac {3}{14}}E_{2}+{\frac {1}{6}}E_{3}+{\frac {9}{88}}E_{2}^{2}-{\frac {3}{22}}E_{4}-{\frac {9}{52}}E_{2}E_{3}+{\frac {3}{26}}E_{5}+O(E_{1})+O(\Delta ^{6})\right)\end{aligned}}}$

As with , expanding around the mean of the arguments eliminates more than half of the terms (those that contain). ${\displaystyle R_{J}}$ ${\displaystyle E_{1}}$

Negative arguments

In general, the arguments , and of Carlson's integrals must not be real and negative, as this would create a branch point on the integration path, which would make the integral ambiguous. However, if the second argument of or the fourth argument of is negative, then there is a simple pole on the integration path. In these cases the principal Cauchy value (the finite part) of the integrals may be of interest; these are ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle z}$ ${\displaystyle R_{C}}$ ${\displaystyle p}$ ${\displaystyle R_{J}}$

${\displaystyle \mathrm {p.v.} \;R_{C}(x,-y)={\sqrt {\frac {x}{x+y}}}\,R_{C}(x+y,y),}$

and

${\displaystyle {\begin{aligned}\mathrm {p.v.} \;R_{J}(x,y,z,-p)&={\frac {(q-y)R_{J}(x,y,z,q)-3R_{F}(x,y,z)+3{\sqrt {y}}R_{C}(xz,-pq)}{y+p}}\\&={\frac {(q-y)R_{J}(x,y,z,q)-3R_{F}(x,y,z)+3{\sqrt {\frac {xyz}{xz+pq}}}R_{C}(xz+pq,pq)}{y+p}}\end{aligned}}}$

in which

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

must be greater than zero so that it can be evaluated. This can be achieved by permuting , and so that the value of lies between that of and . ${\displaystyle R_{J}(x,y,z,q)}$ ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle z}$ ${\displaystyle y}$ ${\displaystyle x}$ ${\displaystyle z}$

Numerical evaluation

The duplication set can be used for a quick and robust evaluation of the symmetrical Carlson forms of elliptical integrals, and thus also for the evaluation of the Legendre form of the elliptical integrals. To compute , we first define , and . Then the series is iterated ${\displaystyle R_{F}(x,y,z)}$ ${\displaystyle x_{0}=x}$ ${\displaystyle y_{0}=y}$ ${\displaystyle z_{0}=z}$

${\displaystyle \lambda _{n}={\sqrt {x_{n}}}{\sqrt {y_{n}}}+{\sqrt {y_{n}}}{\sqrt {z_{n}}}+{\sqrt {z_{n}}}{\sqrt {x_{n}}},}$

${\displaystyle x_{n+1}={\frac {x_{n}+\lambda _{n}}{4}},y_{n+1}={\frac {y_{n}+\lambda _{n}}{4}},z_{n+1}={\frac {z_{n}+\lambda _{n}}{4}}}$

until the desired precision is achieved: If , and are not negative, all series quickly converge to a certain value . So is ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle z}$ ${\displaystyle \mu }$

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

The evaluation of is also done with the help of the relationship ${\displaystyle R_{C}(x,y)}$

${\displaystyle R_{C}\left(x,y\right)=R_{F}\left(x,y,y\right).}$

literature

• BC Carlson, John L. Gustafson: Asymptotic approximations for symmetric elliptic integrals . In: OP-SF 7 . 1993. arxiv : math / 9310223v1 .
• BC Carlson: 'Elliptic Integrals: Symmetric Integrals' in Chap. 19 of Digital Library of Mathematical Functions . National Institute of Standards and Technology. May 7, 2010.
• Fortran code from SLATEC for evaluating RF , RJ , RC , RD ,

Individual evidence

1. ^ 'Profile: Bille C. Carlson' in Digital Library of Mathematical Functions . National Institute of Standards and Technology.
2. ^ Bille C. Carlson: Numerical computation of real or complex elliptic integrals . In: Numerical Algorithms . 1994. arxiv : math / 9409227v1 .
3. ^ WH Press, SA Teukolsky, WT Vetterling, BP Flannery: Section 6.12. Elliptic Integrals and Jacobian Elliptic Functions . In: Numerical Recipes: The Art of Scientific Computing , 3rd. Edition, Cambridge University Press, 2007, ISBN 978-0-521-88068-8 .