functionerfc
Complementary error function erfc(u) = 1 - erf(u)
Extends from Modelica.Icons.Function (Icon for functions).
Information
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
Inputs
| Type | Name | Default | Description |
|---|---|---|---|
| Real | u | Input argument |
Outputs
| Type | Name | Default | Description |
|---|---|---|---|
| Real | y | = 1 - erf(u) |
Revisions
| Date | Description | ||
|---|---|---|---|
| June 22, 2015 |
|