遇到溢出：exp() 乘以 erfc()

Question

I want to evaluate the following formula:

c = exp{x}*erfc{y}

(see definition of x and y in the code below.)

The problem is that x and y are getting quite large and I get very large values for exp{x} and very small values for erfc(y) .

import numpy as np
import scipy as sci
k = 5.7e-3
D = 1.53e-8
R = 1.5e-5
r = 1e-6

t = np.linspace(0.0,12,10)

x = (r/R)  +  (D/(R*R) - k)*t
y = (r/(2*np.sqrt(D*t))) + np.sqrt(D*t)/R

exp_x = np.exp(x)
erfc_y = sci.special.erfc(y)
print("x = \n{} ".format(x))
print("y = \n{}".format(y))

print("exp(x) = \n{}".format(exp_x))
print("erfc(y) = \n{}".format(erfc_y))

print("exp(x) * erfc(y)= \n{}".format(exp_x*erfc_y))

My Idea was to change the evaluation to

log{exp(x)*erfc(y)} = log{exp(x)} + log{erfc(y)} = x + log{erfc(y)}

afterward, I can calculate

exp(x + log{erfc(y)})

But here is the problem: When I want to calculate

log{erfc(y)} = log{1 - erfc(y)} 

I get a similar problem that erfc will be close to 1 and I will get precision problems.

Any ideas to solve my problem?

Answer 1

The solution of this problem was given by Cadena, in this article:

Cadena, F. (1989). Numerical approach to solution of pollutant transport models using personal computers. Computers in Education, 9(2), 34-6.

I could not obtain the article, but the solution is also given by

Lin, JS, & Hildemann, LM (1995). A nonsteady-state analytical model to predict gaseous emissions of volatile organic compounds from landfills. Journal of hazardous materials , 40(3), 271-295.

Cadena proposed to approximate the complementary error function in terms of another exponential function with a negative power term. It reduces the high power raised by the original exponential function that causes overflow.

Suppose W is the multiplication that needs to be evaluated:

W=exp(y)*erfc(x)

W can be approximated as:

W = (a1*t+a2*t^2+a3*t^3+a4*t^4+a5*t^5)*exp(y-x^2), t=1/(1+p*x), if x >= 0

W = 2*exp(y)-(a1*t+a2*t^2+a3*t^3+a4*t^4+a5*t^5)*exp(y-|x|^2), t=1/(1+p*|x|), if x < 0

where

a1 = 0.254829592
a2 = -0.284496736
a3 = 1.42141741
a4 = -1.453152027
a5 = 1.061405429
p = 0.3275911

Answer 2

Use scipy.special.erfcx() which gives exp(x**2)*erfc(x) .

Then you just implement erfcx(x)*exp(xx**2) which again gives erfc(x)*exp(x) .

This of course works best if your arguments x=y .