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Gilles commented on MATH-1386: ------------------------------ bq. When dealing with double I've always used 2.220446049250313E-16 for epsilon. What is the definition of {{epsilon}} in the above? bq. You have it defined in org.apache.commons.math.util.MathUtils as 1.1102230246251565E-16 In recent versions of Commons Math, this constant has been defined in class {{Precision}}. The Commons Math documentation agrees with the following: {noformat} eps = 1.1102230246251565E-16 1 + eps = 1.0 1 + 2 * eps = 1.0000000000000002 1 + nextUp(eps) = 1.0000000000000002 {noformat} See also MATH-843. > EPSILON value in org.apache.commons.math.util.MathUtils seems like half of > what it should be > --------------------------------------------------------------------------------------------- > > Key: MATH-1386 > URL: https://issues.apache.org/jira/browse/MATH-1386 > Project: Commons Math > Issue Type: Bug > Environment: win 7 > Reporter: Brian S. McCormick > Original Estimate: 1h > Remaining Estimate: 1h > > I've always used 1.1920929E-7 for dealing with float epsilon values. When > dealing with double I've always used 2.220446049250313E-16 for epsilon. > You have it defined in org.apache.commons.math.util.MathUtils as > 1.1102230246251565E-16 which is half what I think it should be. > I come up with these numbers using the following: > float fEps = Float.intBitsToFloat(Float.floatToIntBits(1f) + 1) - 1; > double dEps = Double.longBitsToDouble(Double.doubleToLongBits(1) + 1) - 1; > Am I correct? I don't really know. I do know that float epsilon in every > legacy C/C++ compiler etc I have ever used is defined as about 1e-7 and this > is the value using the formula for fEps above. When I started doing comps > using doubles instead of floats I started using the formula for dEps above > which looks to me like the equivalent for double numbers. -- This message was sent by Atlassian JIRA (v6.3.4#6332)