Machine epsilon, Octave and Matlab

[Craig Stoudt]

I have a question about the way Octave handles
arithmetic operations around machine epsilon.

Consider the expression 1/(1-x) - 1/(1+x)
Let x = eps/2.

Since x < eps, when you evaluate this expression in
Octave, the result is zero.  Not surprising.  However,
if you evaluate the same expression in Matlab 6.1, you
get 2.220446049250313e-016 which is eps, which is
pretty close to the correct answer.  If one evaluates
the expression at eps/4, Matlab yields zero, the same
as Octave (the correct answer is approximately eps/2).

I'm just curious about the difference in results
between Matlab and Octave at x = eps/2

Craig Stoudt


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