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.
See also
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