BTW, add this other way of writing a native-precision Newton step to see that it's much worse (numerically) than writing it in the "guess + small_correction" form used in roots.py.  Mathematically they're identical, but numerically they behave differently:


def native2(x, n):
    g = x**(1.0/n)
    if g**n == x:
        return g
    return ((n-1)*g + x/g**(n-1)) / n