This information is part of the Modelica Standard Library maintained by the Modelica Association.

Syntax

Special.erfc(u);

Description

This function computes the complementary error function erfc(u) = 1 - erf(u) with a relative precision of about 1e-15. The implementation utilizes the formulation of the Boost library (53-bit implementation of erf.hpp developed by John Maddock). Plot of the function:

If u is large and erf(u) is subtracted from 1.0, the result is not accurate. It is then better to use erfc(u). For more details, see Wikipedia.

Example

erfc(0)    // = 1
erfc(10)   // = 0
erfc(0.5)  // = 0.4795001221869534

See also

erf, erfInv, erfcInv.