Tim Peters added the comment:
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
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<http://bugs.python.org/issue27761>
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