Serhiy Storchaka added the comment:
What if use pow() with exactly represented degree in approximating step?
def rootn(x, n):
g = x**(1.0/n)
m = 1 << (n-1).bit_length()
if n != m:
g = (x*g**(m-n))**(1.0/m)
return g
Maybe it needs several iterations, because it converges slower than Newton
approximation. But every step can be faster.
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nosy: +serhiy.storchaka
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<https://bugs.python.org/issue27761>
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