Re: a/sqrt(x)

hi, thanks
> The rounding might be different, but it shouldn't be any more in error.
that will be great ..
> If you are that worried about error, maybe you shouldn't be using floating point?
have to use erroneous fp ;.. 
i was looking for a way to equate the results (and so errors) of
possible methods, if the error is not more than normal method that'll
be enough..
how this shud be analyzed ?
thanks again ..