I don't know if this can be considered an improvement over brute-force case analysis: If x is a power of two, then each floating-point operation in x+x+x+x is

I don't know if this can be considered an improvement over brute-force case analysis: If x is a power of two, then each floating-point operation in x+x+x+x is exact. If x isn't a power of two: fl(3x) = 3x + e1 |e1|≤ulp(3x)/2 fl(fl(3x) + x) = fl(3x) + x + e2 |e2|≤ulp(4x)/2 (ulp(4x) because once set apart the case when x is a power of two, there is no way fl(fl(3x) + x) can be in a binade other than the binade of 4x) Now there are two cases: - if 3x is in the binade of 2x, then ulp(3x) = ulp(2x) = ulp(4x)/2. In this case fl(fl(3x) + x) is within (3/4)ulp(4x) of 4x. The only floating-point number within (3/4)ulp(4x) of 4x is 4x. - if 3x is in the binade of 4x: Then the real number 3x cannot be a midpoint for its binade. If floating-point numbers of the binade of 4x are all multiples of 2^p, then all midpoints for this binade are multiples of 2^(p-1). But x is of the form n * 2^(p-2), and 3x = (3n/2) * 2^(p-1) cannot be a midpoint. Since 3x is not a midpoint for its binade, we can in fact improve the inequality |e1|≤ulp(3x)/2 into |e1|<ulp(3x)/2. It follows that fl(fl(3x) + x) is in this case again strictly within ulp(4x) of 4x. I do feel that this is a little hand-wavy as proofs go. In particular, I set apart the case when x is a power of two to make it clear that fl(fl(3x) + x) and 4x are in the same binade in the rest of cases, but I wonder whether it's possible that fl(3x) is a power of two and whether that deserves special attention.