Where do I find the machine epsilon in C#?

Question

The machine epsilon is canonically defined as the smallest number which added to one, gives a result different from one.

There is a Double.Epsilon but t

Answer 1

Just hard-code the value:

const double e1 = 2.2204460492503131e-16;

or use the power of two:

static readonly double e2 = Math.Pow(2, -52);

or use your definition (more or less):

static readonly double e3 = BitConverter.Int64BitsToDouble(BitConverter.DoubleToInt64Bits(1.0) + 1L) - 1.0;

And see Wikipedia: machine epsilon.

Answer 2

It's(on my machine):

   1.11022302462516E-16

You can easily calculate it:

        double machEps = 1.0d;

        do {
           machEps /= 2.0d;
        }
        while ((double)(1.0 + machEps) != 1.0);

        Console.WriteLine( "Calculated machine epsilon: " + machEps );

Edited:

I calcualted 2 times epsilon, now it should be correct.

Answer 3

Ref. the routine in Meonester's: Actually the value of machEps on exit from the do ... while loop is such that 1+machEps == 1. To obtain the machine epsilon we must go back to the previous value, by adding the following after the loop: machEps *= 2.0D; This will return 2.2204460492503131e-16 in agreement with the recommendation in Microsoft's documentation for Double.Epsilon.

Answer 4

LAPACK + DLAMCH, 64-bit INTEL processor, C#:

var pp = double.Epsilon; // pp = 4.94065645841247E-324
double p = NativeMethods.MachinePrecision('S'); // =DLAMCH('S') 
p = 2.2250738585072014E-308
double.MinValue = -1.7976931348623157E+308
double.MaxValue =  1.7976931348623157E+308

Answer 5

The Math.NET library defines a Precision class, which has a DoubleMachineEpsilon property.

You could check how they do it.

According to that it is:

    /// <summary>
    /// The base number for binary values
    /// </summary>
    private const int BinaryBaseNumber = 2;

    /// <summary>
    /// The number of binary digits used to represent the binary number for a double precision floating
    /// point value. i.e. there are this many digits used to represent the
    /// actual number, where in a number as: 0.134556 * 10^5 the digits are 0.134556 and the exponent is 5.
    /// </summary>
    private const int DoublePrecision = 53;

    private static readonly double doubleMachinePrecision = Math.Pow(BinaryBaseNumber, -DoublePrecision);

So it is 1,11022302462516E-16 according to this source.