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:[ 1]
R
F
(
x
,
y
,
z
)
=
1
2
∫
0
∞
d
t
(
t
+
x
)
(
t
+
y
)
(
t
+
z
)
{\displaystyle 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
)
=
3
2
∫
0
∞
d
t
(
t
+
p
)
(
t
+
x
)
(
t
+
y
)
(
t
+
z
)
{\displaystyle 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
)
=
1
4
∫
0
∞
1
(
t
+
x
)
(
t
+
y
)
(
t
+
z
)
(
x
t
+
x
+
y
t
+
y
+
z
t
+
z
)
t
d
t
{\displaystyle 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
)
=
1
2
∫
0
∞
d
t
(
t
+
y
)
(
t
+
x
)
{\displaystyle 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
)
=
3
2
∫
0
∞
d
t
(
t
+
z
)
(
t
+
x
)
(
t
+
y
)
(
t
+
z
)
{\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)}}}}}
Since
R
C
{\displaystyle R_{C}}
and
R
D
{\displaystyle R_{D}}
are special cases of
R
F
{\displaystyle R_{F}}
and
R
J
{\displaystyle R_{J}}
, all elliptic integrals can ultimately be evaluated in terms of just
R
F
{\displaystyle R_{F}}
,
R
J
{\displaystyle R_{J}}
, and
R
G
{\displaystyle 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
)
{\displaystyle R_{F}(x,y,z)}
is the same for any permutation of its arguments, and the value of
R
J
(
x
,
y
,
z
,
p
)
{\displaystyle 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
edit
Incomplete elliptic integrals can be calculated easily using Carlson symmetric forms:
F
(
ϕ
,
k
)
=
sin
ϕ
R
F
(
cos
2
ϕ
,
1
−
k
2
sin
2
ϕ
,
1
)
{\displaystyle F(\phi ,k)=\sin \phi R_{F}\left(\cos ^{2}\phi ,1-k^{2}\sin ^{2}\phi ,1\right)}
E
(
ϕ
,
k
)
=
sin
ϕ
R
F
(
cos
2
ϕ
,
1
−
k
2
sin
2
ϕ
,
1
)
−
1
3
k
2
sin
3
ϕ
R
D
(
cos
2
ϕ
,
1
−
k
2
sin
2
ϕ
,
1
)
{\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)}
Π
(
ϕ
,
n
,
k
)
=
sin
ϕ
R
F
(
cos
2
ϕ
,
1
−
k
2
sin
2
ϕ
,
1
)
+
1
3
n
sin
3
ϕ
R
J
(
cos
2
ϕ
,
1
−
k
2
sin
2
ϕ
,
1
,
1
−
n
sin
2
ϕ
)
{\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)}
(Note: the above are only valid for
−
π
2
≤
ϕ
≤
π
2
{\displaystyle -{\frac {\pi }{2}}\leq \phi \leq {\frac {\pi }{2}}}
and
0
≤
k
2
sin
2
ϕ
≤
1
{\displaystyle 0\leq k^{2}\sin ^{2}\phi \leq 1}
)
Complete elliptic integrals
edit
Complete elliptic integrals can be calculated by substituting φ = 1 ⁄2 π:
K
(
k
)
=
R
F
(
0
,
1
−
k
2
,
1
)
{\displaystyle K(k)=R_{F}\left(0,1-k^{2},1\right)}
E
(
k
)
=
R
F
(
0
,
1
−
k
2
,
1
)
−
1
3
k
2
R
D
(
0
,
1
−
k
2
,
1
)
{\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)}
Π
(
n
,
k
)
=
R
F
(
0
,
1
−
k
2
,
1
)
+
1
3
n
R
J
(
0
,
1
−
k
2
,
1
,
1
−
n
)
{\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)}
When any two, or all three of the arguments of
R
F
{\displaystyle R_{F}}
are the same, then a substitution of
t
+
x
=
u
{\displaystyle {\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
)
=
1
2
∫
0
∞
d
t
t
+
x
(
t
+
y
)
=
∫
x
∞
d
u
u
2
−
x
+
y
=
{
arccos
x
/
y
y
−
x
,
x
<
y
1
y
,
x
=
y
arcosh
x
/
y
x
−
y
,
x
>
y
{\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<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
{\displaystyle R_{J}}
are the same,
R
J
(
x
,
y
,
y
,
p
)
=
3
∫
x
∞
d
u
(
u
2
−
x
+
y
)
(
u
2
−
x
+
p
)
=
{
3
p
−
y
(
R
C
(
x
,
y
)
−
R
C
(
x
,
p
)
)
,
y
≠
p
3
2
(
y
−
x
)
(
R
C
(
x
,
y
)
−
1
y
x
)
,
y
=
p
≠
x
1
y
3
/
2
,
y
=
p
=
x
{\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\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}}}
By substituting in the integral definitions
t
=
κ
u
{\displaystyle t=\kappa u}
for any constant
κ
{\displaystyle \kappa }
, it is found that
R
F
(
κ
x
,
κ
y
,
κ
z
)
=
κ
−
1
/
2
R
F
(
x
,
y
,
z
)
{\displaystyle R_{F}\left(\kappa x,\kappa y,\kappa z\right)=\kappa ^{-1/2}R_{F}(x,y,z)}
R
J
(
κ
x
,
κ
y
,
κ
z
,
κ
p
)
=
κ
−
3
/
2
R
J
(
x
,
y
,
z
,
p
)
{\displaystyle R_{J}\left(\kappa x,\kappa y,\kappa z,\kappa p\right)=\kappa ^{-3/2}R_{J}(x,y,z,p)}
R
F
(
x
,
y
,
z
)
=
2
R
F
(
x
+
λ
,
y
+
λ
,
z
+
λ
)
=
R
F
(
x
+
λ
4
,
y
+
λ
4
,
z
+
λ
4
)
,
{\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),}
where
λ
=
x
y
+
y
z
+
z
x
{\displaystyle \lambda ={\sqrt {x}}{\sqrt {y}}+{\sqrt {y}}{\sqrt {z}}+{\sqrt {z}}{\sqrt {x}}}
.
R
J
(
x
,
y
,
z
,
p
)
=
2
R
J
(
x
+
λ
,
y
+
λ
,
z
+
λ
,
p
+
λ
)
+
6
R
C
(
d
2
,
d
2
+
(
p
−
x
)
(
p
−
y
)
(
p
−
z
)
)
=
1
4
R
J
(
x
+
λ
4
,
y
+
λ
4
,
z
+
λ
4
,
p
+
λ
4
)
+
6
R
C
(
d
2
,
d
2
+
(
p
−
x
)
(
p
−
y
)
(
p
−
z
)
)
{\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}}}
[ 2]
where
d
=
(
p
+
x
)
(
p
+
y
)
(
p
+
z
)
{\displaystyle d=({\sqrt {p}}+{\sqrt {x}})({\sqrt {p}}+{\sqrt {y}})({\sqrt {p}}+{\sqrt {z}})}
and
λ
=
x
y
+
y
z
+
z
x
{\displaystyle \lambda ={\sqrt {x}}{\sqrt {y}}+{\sqrt {y}}{\sqrt {z}}+{\sqrt {z}}{\sqrt {x}}}
In obtaining a Taylor series expansion for
R
F
{\displaystyle R_{F}}
or
R
J
{\displaystyle R_{J}}
it proves convenient to expand about the mean value of the several arguments. So for
R
F
{\displaystyle R_{F}}
, letting the mean value of the arguments be
A
=
(
x
+
y
+
z
)
/
3
{\displaystyle A=(x+y+z)/3}
, and using homogeneity, define
Δ
x
{\displaystyle \Delta x}
,
Δ
y
{\displaystyle \Delta y}
and
Δ
z
{\displaystyle \Delta z}
by
R
F
(
x
,
y
,
z
)
=
R
F
(
A
(
1
−
Δ
x
)
,
A
(
1
−
Δ
y
)
,
A
(
1
−
Δ
z
)
)
=
1
A
R
F
(
1
−
Δ
x
,
1
−
Δ
y
,
1
−
Δ
z
)
{\displaystyle {\begin{aligned}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{aligned}}}
that is
Δ
x
=
1
−
x
/
A
{\displaystyle \Delta x=1-x/A}
etc. The differences
Δ
x
{\displaystyle \Delta x}
,
Δ
y
{\displaystyle \Delta y}
and
Δ
z
{\displaystyle \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
)
{\displaystyle R_{F}(x,y,z)}
is symmetric under permutation of
x
{\displaystyle x}
,
y
{\displaystyle y}
and
z
{\displaystyle z}
, it is also symmetric in the quantities
Δ
x
{\displaystyle \Delta x}
,
Δ
y
{\displaystyle \Delta y}
and
Δ
z
{\displaystyle \Delta z}
. It follows that both the integrand of
R
F
{\displaystyle R_{F}}
and its integral can be expressed as functions of the elementary symmetric polynomials in
Δ
x
{\displaystyle \Delta x}
,
Δ
y
{\displaystyle \Delta y}
and
Δ
z
{\displaystyle \Delta z}
which are
E
1
=
Δ
x
+
Δ
y
+
Δ
z
=
0
{\displaystyle E_{1}=\Delta x+\Delta y+\Delta z=0}
E
2
=
Δ
x
Δ
y
+
Δ
y
Δ
z
+
Δ
z
Δ
x
{\displaystyle E_{2}=\Delta x\Delta y+\Delta y\Delta z+\Delta z\Delta x}
E
3
=
Δ
x
Δ
y
Δ
z
{\displaystyle 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...
R
F
(
x
,
y
,
z
)
=
1
2
A
∫
0
∞
1
(
t
+
1
)
3
−
(
t
+
1
)
2
E
1
+
(
t
+
1
)
E
2
−
E
3
d
t
=
1
2
A
∫
0
∞
(
1
(
t
+
1
)
3
2
−
E
2
2
(
t
+
1
)
7
2
+
E
3
2
(
t
+
1
)
9
2
+
3
E
2
2
8
(
t
+
1
)
11
2
−
3
E
2
E
3
4
(
t
+
1
)
13
2
+
O
(
E
1
)
+
O
(
Δ
6
)
)
d
t
=
1
A
(
1
−
1
10
E
2
+
1
14
E
3
+
1
24
E
2
2
−
3
44
E
2
E
3
+
O
(
E
1
)
+
O
(
Δ
6
)
)
{\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}}}}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 {3E_{2}^{2}}{8(t+1)^{\frac {11}{2}}}}-{\frac {3E_{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{aligned}}}
The advantage of expanding about the mean value of the arguments is now apparent; it reduces
E
1
{\displaystyle E_{1}}
identically to zero, and so eliminates all terms involving
E
1
{\displaystyle E_{1}}
- which otherwise would be the most numerous.
An ascending series for
R
J
{\displaystyle R_{J}}
may be found in a similar way. There is a slight difficulty because
R
J
{\displaystyle R_{J}}
is not fully symmetric; its dependence on its fourth argument,
p
{\displaystyle p}
, is different from its dependence on
x
{\displaystyle x}
,
y
{\displaystyle y}
and
z
{\displaystyle z}
. This is overcome by treating
R
J
{\displaystyle R_{J}}
as a fully symmetric function of five arguments, two of which happen to have the same value
p
{\displaystyle p}
. The mean value of the arguments is therefore taken to be
A
=
x
+
y
+
z
+
2
p
5
{\displaystyle A={\frac {x+y+z+2p}{5}}}
and the differences
Δ
x
{\displaystyle \Delta x}
,
Δ
y
{\displaystyle \Delta y}
Δ
z
{\displaystyle \Delta z}
and
Δ
p
{\displaystyle \Delta p}
defined by
R
J
(
x
,
y
,
z
,
p
)
=
R
J
(
A
(
1
−
Δ
x
)
,
A
(
1
−
Δ
y
)
,
A
(
1
−
Δ
z
)
,
A
(
1
−
Δ
p
)
)
=
1
A
3
2
R
J
(
1
−
Δ
x
,
1
−
Δ
y
,
1
−
Δ
z
,
1
−
Δ
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^{\frac {3}{2}}}}R_{J}(1-\Delta x,1-\Delta y,1-\Delta z,1-\Delta p)\end{aligned}}}
The elementary symmetric polynomials in
Δ
x
{\displaystyle \Delta x}
,
Δ
y
{\displaystyle \Delta y}
,
Δ
z
{\displaystyle \Delta z}
,
Δ
p
{\displaystyle \Delta p}
and (again)
Δ
p
{\displaystyle \Delta p}
are in full
E
1
=
Δ
x
+
Δ
y
+
Δ
z
+
2
Δ
p
=
0
{\displaystyle E_{1}=\Delta x+\Delta y+\Delta z+2\Delta p=0}
E
2
=
Δ
x
Δ
y
+
Δ
y
Δ
z
+
2
Δ
z
Δ
p
+
Δ
p
2
+
2
Δ
p
Δ
x
+
Δ
x
Δ
z
+
2
Δ
y
Δ
p
{\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}
E
3
=
Δ
z
Δ
p
2
+
Δ
x
Δ
p
2
+
2
Δ
x
Δ
y
Δ
p
+
Δ
x
Δ
y
Δ
z
+
2
Δ
y
Δ
z
Δ
p
+
Δ
y
Δ
p
2
+
2
Δ
x
Δ
z
Δ
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}
E
4
=
Δ
y
Δ
z
Δ
p
2
+
Δ
x
Δ
z
Δ
p
2
+
Δ
x
Δ
y
Δ
p
2
+
2
Δ
x
Δ
y
Δ
z
Δ
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}
E
5
=
Δ
x
Δ
y
Δ
z
Δ
p
2
{\displaystyle E_{5}=\Delta x\Delta y\Delta z\Delta p^{2}}
However, it is possible to simplify the formulae for
E
2
{\displaystyle E_{2}}
,
E
3
{\displaystyle E_{3}}
and
E
4
{\displaystyle E_{4}}
using the fact that
E
1
=
0
{\displaystyle E_{1}=0}
. Expressing the integrand in terms of these polynomials, performing a multidimensional Taylor expansion and integrating term-by-term as before...
R
J
(
x
,
y
,
z
,
p
)
=
3
2
A
3
2
∫
0
∞
1
(
t
+
1
)
5
−
(
t
+
1
)
4
E
1
+
(
t
+
1
)
3
E
2
−
(
t
+
1
)
2
E
3
+
(
t
+
1
)
E
4
−
E
5
d
t
=
3
2
A
3
2
∫
0
∞
(
1
(
t
+
1
)
5
2
−
E
2
2
(
t
+
1
)
9
2
+
E
3
2
(
t
+
1
)
11
2
+
3
E
2
2
−
4
E
4
8
(
t
+
1
)
13
2
+
2
E
5
−
3
E
2
E
3
4
(
t
+
1
)
15
2
+
O
(
E
1
)
+
O
(
Δ
6
)
)
d
t
=
1
A
3
2
(
1
−
3
14
E
2
+
1
6
E
3
+
9
88
E
2
2
−
3
22
E
4
−
9
52
E
2
E
3
+
3
26
E
5
+
O
(
E
1
)
+
O
(
Δ
6
)
)
{\displaystyle {\begin{aligned}R_{J}(x,y,z,p)&={\frac {3}{2A^{\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}{2A^{\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 {3E_{2}^{2}-4E_{4}}{8(t+1)^{\frac {13}{2}}}}+{\frac {2E_{5}-3E_{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{aligned}}}
As with
R
J
{\displaystyle R_{J}}
, by expanding about the mean value of the arguments, more than half the terms (those involving
E
1
{\displaystyle E_{1}}
) are eliminated.
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
{\displaystyle R_{C}}
, or the fourth argument, p, of
R
J
{\displaystyle 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
p
.
v
.
R
C
(
x
,
−
y
)
=
x
x
+
y
R
C
(
x
+
y
,
y
)
,
{\displaystyle \mathrm {p.v.} \;R_{C}(x,-y)={\sqrt {\frac {x}{x+y}}}\,R_{C}(x+y,y),}
and
p
.
v
.
R
J
(
x
,
y
,
z
,
−
p
)
=
(
q
−
y
)
R
J
(
x
,
y
,
z
,
q
)
−
3
R
F
(
x
,
y
,
z
)
+
3
y
R
C
(
x
z
,
−
p
q
)
y
+
p
=
(
q
−
y
)
R
J
(
x
,
y
,
z
,
q
)
−
3
R
F
(
x
,
y
,
z
)
+
3
x
y
z
x
z
+
p
q
R
C
(
x
z
+
p
q
,
p
q
)
y
+
p
{\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}}}
where
q
=
y
+
(
z
−
y
)
(
y
−
x
)
y
+
p
.
{\displaystyle q=y+{\frac {(z-y)(y-x)}{y+p}}.}
which must be greater than zero for
R
J
(
x
,
y
,
z
,
q
)
{\displaystyle 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
edit
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
)
{\displaystyle R_{F}(x,y,z)}
:
first, define
x
0
=
x
{\displaystyle x_{0}=x}
,
y
0
=
y
{\displaystyle y_{0}=y}
and
z
0
=
z
{\displaystyle z_{0}=z}
. Then iterate the series
λ
n
=
x
n
y
n
+
y
n
z
n
+
z
n
x
n
,
{\displaystyle \lambda _{n}={\sqrt {x_{n}}}{\sqrt {y_{n}}}+{\sqrt {y_{n}}}{\sqrt {z_{n}}}+{\sqrt {z_{n}}}{\sqrt {x_{n}}},}
x
n
+
1
=
x
n
+
λ
n
4
,
y
n
+
1
=
y
n
+
λ
n
4
,
z
n
+
1
=
z
n
+
λ
n
4
{\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 reached: if
x
{\displaystyle x}
,
y
{\displaystyle y}
and
z
{\displaystyle z}
are non-negative, all of the series will converge quickly to a given value, say,
μ
{\displaystyle \mu }
. Therefore,
R
F
(
x
,
y
,
z
)
=
R
F
(
μ
,
μ
,
μ
)
=
μ
−
1
/
2
.
{\displaystyle R_{F}\left(x,y,z\right)=R_{F}\left(\mu ,\mu ,\mu \right)=\mu ^{-1/2}.}
Evaluating
R
C
(
x
,
y
)
{\displaystyle R_{C}(x,y)}
is much the same due to the relation
R
C
(
x
,
y
)
=
R
F
(
x
,
y
,
y
)
.
{\displaystyle R_{C}\left(x,y\right)=R_{F}\left(x,y,y\right).}
References and External links
edit
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 , archived from the original on 2011-08-11, retrieved 2011-08-10
Fortran code from SLATEC for evaluating RF , RJ , RC , RD ,