Mark Dickinson added the comment: > Mark, the code I showed in roots.py is somewhat more accurate and highly > significantly faster than the code you just posted.
Okay, fair enough. In that case, we still need a solution for computing rootn(x * 2**e) in the case where x*2**e itself overflows / underflows an IEEE 754 float. ---------- _______________________________________ Python tracker <rep...@bugs.python.org> <http://bugs.python.org/issue27761> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
