Performance of norm function

General Usage Performance
21

To calculate that correctly, initialize x and y as BigFloats using strings.

julia> x = BigFloat("3e-200"); y = BigFloat("4e-200");
julia> sqrt(x^2 + y^2) 
4.99999999999999999999999999999999999999999999999999999999999999999999999997e-200

otherwise the erosion has already occured when the conversion to BigFloat happens

1 Like
22

Yes, I was responding to “All it would require is doing the accumulation (not the squaring) in twice precision.” If efficiency is the concern, better to use the n-d generalization of the hypot algorithm |x| \sqrt{1 +(y/x)^2} in the original precision, where x is the largest element of the vector being summed.

3 Likes