- Carlson symmetric form
In
mathematics , the Carlson symmetric forms ofelliptic integral s, R_C, R_D, R_F and R_J are defined by:R_C(x,y) := frac{1}{2} int_0^infty (t+x)^{-1/2} (t+y)^{-1},dt
:R_D(x,y,z) := frac{3}{2} int_0^infty (t+x)^{-1/2} (t+y)^{-1/2} (t+z)^{-3/2},dt
:R_F(x,y,z) := frac{1}{2} int_0^infty (t+x)^{-1/2} (t+y)^{-1/2} (t+z)^{-1/2},dt
:R_J(x,y,z,p) := frac{3}{2} int_0^infty (t+x)^{-1/2} (t+y)^{-1/2} (t+z)^{-1/2} (t+p)^{-1},dt
Note that R_C is a special case of R_F and R_D is a special case of R_J;
:R_Cleft(x,y ight)=R_Fleft(x,y,y ight)
:R_Dleft(x,y,z ight)=R_Jleft(x,y,z,z ight).
The term "symmetric" refers to the fact that these functions are unchanged by the exchange of certain of their arguments. The value of R_F(x,y,z) is the same for any permutation of its arguments, and the value of R_J(x,y,z,p) is the same for any permutation of its first three arguments.
Relations concerning to
Legendre form ofelliptic integral sIncomplete elliptic integrals
Incomplete
elliptic integral s can be calculated easily using Carlson symmetric forms::F(phi,k)=sinphi R_Fleft(cos^2phi,1-k^2sin^2phi,1 ight)
:E(phi,k)=sinphi R_Fleft(cos^2phi,1-k^2sin^2phi,1 ight) -frac{1}{3}k^2sin^3phi R_Dleft(cos^2phi,1-k^2sin^2phi,1 ight)
:Pi(phi,n,k)=sinphi R_Fleft(cos^2phi,1-k^2sin^2phi,1 ight)+frac{1}{3}nsin^3phi R_Jleft(cos^2phi,1-k^2sin^2phi,1,1-nsin^2phi ight)
Complete elliptic integrals
Complete
elliptic integral s can be calculated by substituting phi=frac{pi}{2}::K(k)=R_Fleft(0,1-k^2,1 ight)
:E(k)=R_Fleft(0,1-k^2,1 ight)-frac{1}{3}k^2 R_Dleft(0,1-k^2,1 ight)
:Pi(n,k)=R_Fleft(0,1-k^2,1 ight)+frac{1}{3}n R_J left(0,1-k^2,1,1-n ight)
pecial 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 =egin{cases} frac{arccosleft( sqrt{x/y} ight)}{sqrt{y - x, & x < y \ frac{1}{sqrt{y, & x = y \ frac{mathrm{arccosh}left( sqrt{x/y} ight)}{sqrt{x - y, & x > y \ end{cases}
Other properties
Homogeneity
By substituting in the integral definitions t = kappa u for any constant kappa, it is found that
:R_Fleft(kappa x,kappa y,kappa z ight)=kappa^{-1/2}R_F(x,y,z)
:R_Jleft(kappa x,kappa y,kappa z,kappa p ight)=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_Fleft(frac{x+lambda}{4},frac{y+lambda}{4},frac{z+lambda}{4} ight),
where lambda=sqrt{xy}+sqrt{yz}+sqrt{zx}.
:egin{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} ight) + 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}
Numerical evaluation
The duplication theorem can be used for a fast and robust evaluation of the Carlson symmetric form of elliptic integralsand therefore also for the evaluation of Legendre-form of elliptic integrals. Let us calculate R_F(x,y,z):first, define x_0=x, y_0=y and z_0=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, mu. Therefore,
:R_Fleft(x,y,z ight)=R_Fleft(mu,mu,mu ight)=mu^{-1/2}.
Evaluating R_C(x,y) is much the same due to the relation
:R_Cleft(x,y ight)=R_Fleft(x,y,y ight).
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]
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