2025年2月27日 · Erfc is the complementary error function, commonly denoted erfc(z), is an entire function defined by erfc(z) = 1-erf(z) (1) = 2/(sqrt(pi))int_z^inftye^(-t^2)dt. (2) It is implemented in the Wolfram Language as Erfc[z].

ERFC (x) returns the error function integrated between x and infinity.

erfc(x) returns the Complementary Error Function evaluated for each element of x. Use the erfc function to replace 1 - erf(x) for greater accuracy when erf(x) is close to 1.

2012年7月16日 · The Q-function gives the probability that a random variable from a normal distribution will exceed a certain threshold value. The erf function gives the probability that a normally distributed variable will fall within a certain range.

The complementary error function erfc(x) is plotted in Fig. 2, along with an upper and a lower bound (established in Appendix A). The bounds are asymptotically tight, i.e., the di erence between the bound and the actual function converges to zero for large x.

Mathematical function, suitable for both symbolic and numerical manipulation. Erfc [ z ] is given by . For certain special arguments, Erfc automatically evaluates to exact values.

The value of erf(z) can be found in mathematical tables, as build-in functions in calculators and spread sheets. If you have a programmable calculator, you may find the following approximation useful (it is accurate to 1 part in 107): erf(z) = 1 - (a. 1T + a2T. 2 +a3T3 +a4T4 +a5T 5) e-z 2.

Erfc[z] (235 formulas) Primary definition (1 formula) Specific values (6 formulas) General characteristics (5 formulas) Series representations (17 formulas) Integral representations (5 formulas) Continued fraction representations (11 formulas) Differential equations (11 formulas) Transformations (3 formulas) Complex characteristics (10 formulas)

The complementary error function, denoted as 'erfc', is a mathematical function that describes the probability of a random variable falling outside a specified range for normally distributed data. …

Function (noted erfc) is defined by the following second order differential equation 2x ∂y(x) ∂x + ∂2y(x) ∂x2 (ERFC.1.1) = 0. The initial conditions of ERFC.1.1 are given at 0 by erfc(0) = 1, …