# scipy.special.elliprc#

scipy.special.elliprc(x, y, out=None) = <ufunc 'elliprc'>#

Degenerate symmetric elliptic integral.

The function RC is defined as [1]

$R_{\mathrm{C}}(x, y) = \frac{1}{2} \int_0^{+\infty} (t + x)^{-1/2} (t + y)^{-1} dt = R_{\mathrm{F}}(x, y, y)$
Parameters
x, yarray_like

Real or complex input parameters. x can be any number in the complex plane cut along the negative real axis. y must be non-zero.

outndarray, optional

Optional output array for the function values

Returns
Rscalar or ndarray

Value of the integral. If y is real and negative, the Cauchy principal value is returned. If both of x and y are real, the return value is real. Otherwise, the return value is complex.

elliprf

Completely-symmetric elliptic integral of the first kind.

elliprd

Symmetric elliptic integral of the second kind.

elliprg

Completely-symmetric elliptic integral of the second kind.

elliprj

Symmetric elliptic integral of the third kind.

Notes

RC is a degenerate case of the symmetric integral RF: elliprc(x, y) == elliprf(x, y, y). It is an elementary function rather than an elliptic integral.

The code implements Carlson’s algorithm based on the duplication theorems and series expansion up to the 7th order. [2]

New in version 1.8.0.

References

1

B. C. Carlson, ed., Chapter 19 in “Digital Library of Mathematical Functions,” NIST, US Dept. of Commerce. https://dlmf.nist.gov/19.16.E6

2

B. C. Carlson, “Numerical computation of real or complex elliptic integrals,” Numer. Algorithm, vol. 10, no. 1, pp. 13-26, 1995. https://arxiv.org/abs/math/9409227 https://doi.org/10.1007/BF02198293

Examples

Basic homogeneity property:

>>> from scipy.special import elliprc

>>> x = 1.2 + 3.4j
>>> y = 5.
>>> scale = 0.3 + 0.4j
>>> elliprc(scale*x, scale*y)
(0.5484493976710874-0.4169557678995833j)

>>> elliprc(x, y)/np.sqrt(scale)
(0.5484493976710874-0.41695576789958333j)


When the two arguments coincide, the integral is particularly simple:

>>> x = 1.2 + 3.4j
>>> elliprc(x, x)
(0.4299173120614631-0.3041729818745595j)

>>> 1/np.sqrt(x)
(0.4299173120614631-0.30417298187455954j)


Another simple case: the first argument vanishes:

>>> y = 1.2 + 3.4j
>>> elliprc(0, y)
(0.6753125346116815-0.47779380263880866j)

>>> np.pi/2/np.sqrt(y)
(0.6753125346116815-0.4777938026388088j)


When x and y are both positive, we can express $$R_C(x,y)$$ in terms of more elementary functions. For the case $$0 \le x < y$$,

>>> x = 3.2
>>> y = 6.
>>> elliprc(x, y)
0.44942991498453444

>>> np.arctan(np.sqrt((y-x)/x))/np.sqrt(y-x)
0.44942991498453433


And for the case $$0 \le y < x$$,

>>> x = 6.
>>> y = 3.2
>>> elliprc(x,y)
0.4989837501576147

>>> np.log((np.sqrt(x)+np.sqrt(x-y))/np.sqrt(y))/np.sqrt(x-y)
0.49898375015761476